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Using definite integrals to find area and length

Early on in our work with the definite integral, we learned that for an object moving along an axis, the area under a non-negative velocity function v between a and b tells us the distance the object traveled on that…

Introduction

Early on in our work with the definite integral, we learned that for an object moving along an axis, the area under a non-negative velocity function \(v\) between \(a\) and \(b\) tells us the distance the object traveled on that time interval, and that area is given precisely by the definite integral \(\int_a^b v(t) \, dt\). In general, for any nonnegative function \(f\) on an interval \([a,b]\), \(\int_a^b f(x) \, dx\) measures the area bounded by the curve and the \(x\)-axis between \(x = a\) and \(x = b\).

Next, we will explore how definite integrals can be used to represent other physically important properties. In Preview Activity, we investigate how a single definite integral may be used to represent the area between two curves.

Exploration

The Area Between Two Curves

In Preview Activity, we saw a natural way to think about the area between two curves: it is the area beneath the upper curve minus the area below the lower curve.

Example

Find the area bounded between the graphs of \(f(x) = (x-1)^2 + 1\) and \(g(x) = x+2\).

Solution

We can find the intersection points of the graphs of \(f\) and \(g\) algebraically by solving the system of equations given by \(y = x+2\) and \(y = (x-1)^2 + 1\): substituting \(x+2\) for \(y\) in the second equation yields \(x+2 = (x-1)^2 + 1\), so \(x+2 = x^2 - 2x + 1 + 1\), and thus \[\begin{aligned}\end{aligned}\], from which it follows that \(x = 0\) or \(x = 3\). Using \(y = x+2\), we find the corresponding \(y\)-values of the intersection points. In Figure, we confirm this result and show the \(y\)-coordinates of the points, \((0,2)\) and \((3,5)\). Furthermore, the figure shows how we can think of the area beneath \(f\) and the area beneath \(g\) separately, and use the difference of these quantities to find the difference between the curves.

Now, using definite integrals to find the values of the respective areas beneath the curves, on the interval \([0,3]\), the area beneath \(g\) is \[\begin{aligned}\end{aligned}\], while the area under \(f\) on the same interval is \[\begin{aligned}\end{aligned}\].

Thus, the area between the curves is \[\begin{aligned}\end{aligned}\].

We can also think of the area in this slightly different way: if we slice up the region between two curves into thin vertical rectangles (in the same spirit as we originally sliced the region between a single curve and the \(x\)-axis in Section), we see (as shown in Figure) that the height of a typical rectangle is given by the difference between the two functions, \(g(x) - f(x)\), and its width is \(\Delta x\). Thus the area of the rectangle is \[\begin{aligned}\end{aligned}\].

The area between the two curves on \([0,3]\) is thus approximated by the Riemann sum \[\begin{aligned}\end{aligned}\], and as we let \(n \to \infty\), it follows that the area is given by the single definite integral \[\begin{aligned}\end{aligned}\].

In many applications of the definite integral, we will find it helpful to think of a representative slice and use the definite integral to add up these slices. Here, the integral sums the areas of thin rectangles.

Our work so far in this section illustrates the following general principle.

Condensed — the full section is in Boelkins, Active Calculus.

Finding Area with Horizontal Slices

At times, the shape of a region may dictate that we use horizontal rectangular slices, instead of vertical ones.

Example

Find the area of the region bounded by the parabola \(x = y^2 - 1\) and the line \(y = x-1\), shown at left in Figure.

Solution

By solving the second equation for \(x\) and writing \(x = y + 1\), we find that \(y+1 = y^2 - 1\). Hence the curves intersect where \(y^2 - y - 2 = 0\). Thus, we find \(y = -1\) or \(y = 2\), so the intersection points of the two curves are \((0,-1)\) and \((3,2)\).

If we attempt to use vertical rectangles to slice up the area (as in the center graph of Figure), we see that from \(x = -1\) to \(x = 0\) the curves that bound the top and bottom of the rectangle are one and the same. This suggests, as shown in the rightmost graph in the figure, that we try using horizontal rectangles.

Note that the width of a horizontal rectangle depends on \(y\). Between \(y = -1\) and \(y = 2\), the right end of a representative rectangle is determined by the line \(x = y+1\), and the left end is determined by the parabola, \(x = y^2-1\). The thickness of the rectangle is \(\Delta y\).

Therefore, the area of the rectangle is \[\begin{aligned}\end{aligned}\], and the area between the two curves on the \(y\)-interval \([-1,2]\) is approximated by the Riemann sum \[\begin{aligned}\end{aligned}\].

Taking the limit of the Riemann sum, it follows that the area of the region is \[\begin{aligned}\end{aligned}\].

We emphasize that we are integrating with respect to \(y\); this is because we chose to use horizontal rectangles whose widths depend on \(y\) and whose thickness is denoted \(\Delta y\). It is a straightforward exercise to evaluate the integral in Equation and find that \(A = \frac{9}{2}\).

Just as with the use of vertical rectangles of thickness \(\Delta x\), we have a general principle for finding the area between two curves, which we state as follows.

If two curves \(x = g(y)\) and \(x = f(y)\) intersect at \((g(c),c)\) and \((g(d),d)\), and for all \(y\) such that \(c \le y \le d\), \(g(y) \ge f(y)\), then the area between the curves is \[\begin{aligned}\end{aligned}\].

Finding the length of a curve

We can also use the definite integral to find the length of a portion of a curve. We use the same fundamental principle: we slice the curve up into small pieces whose lengths we can easily approximate. Specifically, we subdivide the curve into small approximating line segments, as shown at left in Figure.

We estimate the length \(L_{\text{slice} }\) of each portion of the curve on a small interval of length \(\Delta x\). We use the right triangle with legs parallel to the coordinate axes and hypotenuse connecting the endpoints of the slice, as seen at right in Figure. The length, \(h\), of the hypotenuse approximates the length, \(L_{\text{slice} }\), of the curve between the two selected points. Thus, \[\begin{aligned}\end{aligned}\].

Next we use algebra to rearrange the expression for the length of the hypotenuse into a form that we can integrate. By removing a factor of \((\Delta x)^2\), we find \[\begin{aligned}L_{\text{slice}} &\approx \sqrt{ (\Delta x)^2 + (\Delta y)^2 } \\ &= \sqrt{ (\Delta x)^2\left(1 + \frac{(\Delta y)^2}{(\Delta x)^2} \right)} \\ &= \sqrt{1 + \frac{(\Delta y)^2}{(\Delta x)^2} } \cdot \Delta x\end{aligned}\].

Then, as \(n \to \infty\) and \(\Delta x \to 0\), we have that \(\frac{\Delta y}{\Delta x} \to \frac{dy}{dx} = f'(x)\). Thus, we can say that \[\begin{aligned}\end{aligned}\].

Taking a Riemann sum of all of these slices and letting \(n \to \infty\), the total length of the curve is therefore \[\begin{aligned}\end{aligned}\].

We summarize this result more formally as follows.

Given a differentiable function \(f\) on an interval \([a,b]\), the total arc length, \(L\), along the curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by \[\begin{aligned}\end{aligned}\].

Summary

To find the area between two curves, we think about slicing the region into thin rectangles. If, for instance, the area of a typical rectangle on the interval \(x = a\) to \(x = b\) is given by \(A_{\text{rect} } = (g(x) - f(x)) \Delta x\), then the exact area of the region is given by the definite integral \[\begin{aligned}\end{aligned}\].
The shape of the region usually dictates whether we should use vertical rectangles of thickness \(\Delta x\) or horizontal rectangles of thickness \(\Delta y\). We want the height of the rectangle given by the difference between two curves: if those curves are best thought of as functions of \(y\), we use horizontal rectangles, whereas if those curves are best viewed as functions of \(x\), we use vertical rectangles.
The arc length, \(L\), along the curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by \[\begin{aligned}\end{aligned}\].

Practice (6)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the exact area of each described region.
1. The finite region between the curves \(x = y(y-2)\) and \(x=-(y-1)(y-3)\).
2. The region between the sine and cosine functions on the interval \([\frac{\pi}{4}, \frac{3\pi}{4}]\).
3. The finite region between \(x = y^2 - y - 2\) and \(y = 2x-1\).
4. The finite region between \(y = mx\) and \(y = x^2-1\), where \(m\) is a positive constant.
Αποκάλυψέ την.
1. The point of intersection of the graphs occurs when \(y(y - 2) = -(y - 1)(y - 3)\), and thus we need to find where \(y^2 - 2y = -(y^2 - 4y + 3)\) or \(2y^2 - 6y + 3 = 0\). By the quadratic formula, it follows that \[\begin{aligned}\end{aligned}\]. Since the curve \(x = -(y - 1)(y - 3)\) lies to the right (has a greater \(x\) value) of \(x = y(y - 2)\) between \(y = \frac{3 - \sqrt{3}}{2} \approx 0.63\) and \(y = \frac{3 + \sqrt{3}}{2} \approx 2.37\) as shown in the figure below, we see that the area between the two curves is given by the integral \[\begin{aligned}\end{aligned}\]. Evaluating this integral, we find that \[\begin{aligned}A &= \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -2y^2 + 6y - 3 \ dy \\ &= \left. \left(-\frac{2}{3}y^3 + 3y^2 - 3y \right) \right|_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} \\ &= \left[-\frac{2}{3} \left(\frac{3 + \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 + \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 + \sqrt{3}}{2}\right)\right] \\ &\qquad - \left[-\frac{2}{3} \left(\frac{3 - \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 - \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 - \sqrt{3}}{2}\right)\right] \\ &= -\frac{1}{12}(54+30\sqrt{3}) + \frac{1}{12}(54-30\sqrt{3}) + \frac{3}{4}(12+6\sqrt{3}) \\ &\qquad - \frac{3}{4}(12-6\sqrt{3}) - \frac{3}{2}(3+\sqrt{3}) + \frac{3}{2}(3-\sqrt{3}) \\ &= -5\sqrt{3} + 9\sqrt{3} - 3\sqrt{3} \\ &= \sqrt{3}\end{aligned}\].
2. As the figure below shows, the graph of \(y = \sin(x)\) lies above the graph of \(y = \cos(x)\) on this interval. So the area between the two curves is given by \[\begin{aligned}A &= \int_{\pi/4}^{3\pi/4} \sin(x) - \cos(x) \ dx \\ &= \left. \left(-\cos(x) - \sin(x) \right) \right|_{\pi/4}^{3\pi/4} \\ &= \left[-\cos\left(\frac{3\pi}{4}\right) - \sin\left(\frac{3\pi}{4}\right) \right] - \left[-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) \right] \\ &= \left[ \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right] + \left[ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right] \\ &= \sqrt{2}\end{aligned}\].
3. We can rewrite \(y = 2x - 1\) as \(x = \frac{y+1}{2}\). The two curves intersect when \(y^2 - y - 2 = \frac{y+1}{2}\). It follows that \(2y^2 - 2y - 4 = y+1\), so \(2y^2 - 3y - 5 = 0\) and thus factoring, \((2y-5)(y+1) = 0\). Hence the two curves intersect when \(y = -1\) or \(y = \frac{5}{2}\). The figure below shows that \(x = \frac{y+1}{2}\) lies to the right of \(x = y^2 - y - 2\)on this interval, so the area of the region between the two curves is \[\begin{aligned}A &= \int_{-1}{5/2} \frac{y+1}{2} - (y^2-y-2) \ dy \\ &= \int_{-1}{5/2} -y^2 + \frac{3}{2}y + \frac{5}{2} \ dy \\ &= \left. \left( -\frac{y^3}{3} + \frac{3}{4}y^2 + \frac{5}{2}y \right) \right|_{-1}{5/2} \\ &= -\frac{1}{3}\left[\left(\frac{5}{2}\right)^3 + 1 \right] + \frac{3}{4}\left[\left(\frac{5}{2}\right)^2 - 1 \right] + \frac{5}{2}\left[\left(\frac{5}{2}\right) + 1 \right] \\ &= -\frac{133}{24} + \frac{63}{16} + \frac{35}{4} \\ &= \frac{343}{48}\end{aligned}\].
4. A representative picture (with \(m = 2\)) is shown in the figure below. The two curves intersect when \(x^2 - 1 = mx\), or when \(x^2 - mx - 1 = 0\). By the quadratic formula, it folows that \[\begin{aligned}\end{aligned}\]. If we integrate with vertical slices, then the graph of \(y = mx\) is always above the graph of \(y = x^2 - 1\) and so the area of the region between the two graphs is \[\begin{aligned}A &= \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx \\ &= \left. \left(-\frac{x^3}{3} + m\frac{x^2}{2} + x\right) \right|_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} \\ &= - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2 \\ &\qquad - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right)\end{aligned}\].
Let \(f(x) = 1-x^2\) and \(g(x) = ax^2 - a\), where \(a\) is an unknown positive real number. For what value(s) of \(a\) is the area between the curves \(f\) and \(g\) equal to 2?

Αποκάλυψέ την.

Both of these quadratic functions pass through the points \((-1,0)\) and \((1,0)\), and thus these are the two intersection points for all positive values of \(a\). In addition, since \(1 - x^2 \gt ax^2 - a\) on \(-1 \lt x \lt 1\), it follows that the area between the curves is given by \[\begin{aligned}\end{aligned}\]. Evaluating the integral, we find that \[\begin{aligned}A &= \int_{-1}^{1} \left( 1 + a - (1+a)x^2 \right) \, dx \\ &= \left. (1+a)x - \frac{1+a}{3}x^3 \right|_{-1}^{1} \\ &= (1+a)(1) - \frac{1+a}{3}(1)^3 - \left( (1+a)(-1) - \frac{1+a}{3}(-1)^3 \right) \\ &= 2(1+a) - \frac{2(1+a)}{3} \\ &= \frac{4(1+a)}{3}\end{aligned}\]. We want to know for which value(s) of \(a\) it follows that \(A = 2\). Thus, we solve \[\begin{aligned}\end{aligned}\], from which it follows that \(a = \frac{1}{2}\). Hence, in the figure below, the area between the two curves is exactly \(A = 2\).
Let \(f(x) = 2-x^2\). Recall that the average value of any continuous function \(f\) on an interval \([a,b]\) is given by \(\frac{1}{b-a} \int_a^b f(x) \, dx\).
1. Find the average value of \(f(x) = 2-x^2\) on the interval \([0,\sqrt{2}]\). Call this value \(r\).
2. Sketch a graph of \(y = f(x)\) and \(y = r\). Find their intersection point(s).
3. Show that on the interval \([0,\sqrt{2}]\), the amount of area that lies below \(y = f(x)\) and above \(y = r\) is equal to the amount of area that lies below \(y = r\) and above \(y = f(x)\).
4. Will the result of (c) be true for any continuous function and its average value on any interval? Why?
Αποκάλυψέ την.
1. The average value of \(f\) on the interval \([0,\sqrt{2}]\) is \[\begin{aligned}r &= \frac{1}{\sqrt{2}} \int_0^{\sqrt{2}} 2-x^2 \, dx \\ &= \left. \frac{1}{\sqrt{2}}\left(2x - \frac{x^3}{3}\right) \right|_0^{\sqrt{2}} \\ &= \frac{1}{\sqrt{2}}\left( 2\sqrt{2} - \frac{\sqrt{2}^3}{3} \right) \\ &= 2 - \frac{2}{3} \\ &= \frac{4}{3}\end{aligned}\].
2. The intersection point of \(y=f(x)\) and \(y=r\) occurs, for positive \(x\), when \(2-x^2 = \frac{4}{3}\), from which it follows that \(x^2 = \frac{2}{3}\) and thus \(x = \frac{\sqrt{6}}{3}\). A graph of \(f\), \(y = r\), and the intersection point is shown in the following figure.
3. Let \(A_1\) be the area that lies below \(y = f(x)\) and above \(y = r\) on the interval \(\left[0, \frac{\sqrt{6}}{3}\right]\) and \(A_2\) the are of the region that lies below \(y = r\) and above \(y = f(x)\) on \(\left[\frac{\sqrt{6}}{3}, \sqrt{2}\right]\) as illustrated in the figure below. Then \[\begin{aligned}A_1 &= \int_0^{\sqrt{6}/3} (2-x^2) - \frac{4}{3} \ dx \\ &= \int_0^{\sqrt{6}/3} \frac{2}{3} - x^2 \ dx \\ &= \left. \left( \frac{2}{3}x - \frac{x^3}{3} \right) \right|_0^{\sqrt{6}/3} \\ &= \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) - \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 \\ &= \frac{2\sqrt{6}}{9} - \frac{6 \sqrt{6}}{81} \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\]. and \[\begin{aligned}A_2 &= \int_{\sqrt{6}/3}^{\sqrt{2}} \frac{4}{3} - (2-x^2) \ dx \\ &= \int_{\sqrt{6}/3}^{\sqrt{2}} x^2 - \frac{2}{3} \ dx \\ &= \left. \left(\frac{x^3}{3} - \frac{2}{3}x \right) \right|_{\sqrt{6}/3}^{\sqrt{2}} \\ &= \left(\frac{2\sqrt{2}}{3} - \frac{2}{3} \sqrt{2} \right) - \left[ \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 - \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) \right] \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\]. Hence \(A_1 = A_2\) as desired.
4. The answer is yes. Let \(AV\) is the average value of a function \(f\) on an interval \([a,b]\). By the definition of average value, \[\begin{aligned}\end{aligned}\], which can also be viewed as the area of the rectangle with base \((b-a)\) and height \(AV\). Now suppose that the area of the region above the graph of \(f\) and below the line \(y=AV\) on the interval \([a,b]\) is \(R_1\) and the area of the region below the graph of \(f\) and above the line \(y=AV\) on the interval \([a,b]\) is \(R_2\). The figure below illustrates that the value of the definite integral is the area of the rectangle plus the area above the rectangle minus the area below the rectangle, or \[\begin{aligned}\end{aligned}\]. Having already established that \(AV(b-a) = \int_a^b f(x) \ dx\), it follows that \(R_2 - R_1 = 0\). So the area of the region that lies below \(y = f(x)\) and above \(y = AV\) is equal to the amount of area that lies below \(y = AV\) and above \(y = f(x)\) on \([a,b]\).
Find the exact area of each described region.
1. The finite region between the curves \(x = y(y-2)\) and \(x=-(y-1)(y-3)\).
2. The region between the sine and cosine functions on the interval \([\frac{\pi}{4}, \frac{3\pi}{4}]\).
3. The finite region between \(x = y^2 - y - 2\) and \(y = 2x-1\).
4. The finite region between \(y = mx\) and \(y = x^2-1\), where \(m\) is a positive constant.
Αποκάλυψέ την.
1. The point of intersection of the graphs occurs when \(y(y - 2) = -(y - 1)(y - 3)\), and thus we need to find where \(y^2 - 2y = -(y^2 - 4y + 3)\) or \(2y^2 - 6y + 3 = 0\). By the quadratic formula, it follows that \[\begin{aligned}\end{aligned}\]. Since the curve \(x = -(y - 1)(y - 3)\) lies to the right (has a greater \(x\) value) of \(x = y(y - 2)\) between \(y = \frac{3 - \sqrt{3}}{2} \approx 0.63\) and \(y = \frac{3 + \sqrt{3}}{2} \approx 2.37\) as shown in the figure below, we see that the area between the two curves is given by the integral \[\begin{aligned}\end{aligned}\]. Evaluating this integral, we find that \[\begin{aligned}A &= \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -2y^2 + 6y - 3 \ dy \\ &= \left. \left(-\frac{2}{3}y^3 + 3y^2 - 3y \right) \right|_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} \\ &= \left[-\frac{2}{3} \left(\frac{3 + \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 + \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 + \sqrt{3}}{2}\right)\right] \\ &\qquad - \left[-\frac{2}{3} \left(\frac{3 - \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 - \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 - \sqrt{3}}{2}\right)\right] \\ &= -\frac{1}{12}(54+30\sqrt{3}) + \frac{1}{12}(54-30\sqrt{3}) + \frac{3}{4}(12+6\sqrt{3}) \\ &\qquad - \frac{3}{4}(12-6\sqrt{3}) - \frac{3}{2}(3+\sqrt{3}) + \frac{3}{2}(3-\sqrt{3}) \\ &= -5\sqrt{3} + 9\sqrt{3} - 3\sqrt{3} \\ &= \sqrt{3}\end{aligned}\].
2. As the figure below shows, the graph of \(y = \sin(x)\) lies above the graph of \(y = \cos(x)\) on this interval. So the area between the two curves is given by \[\begin{aligned}A &= \int_{\pi/4}^{3\pi/4} \sin(x) - \cos(x) \ dx \\ &= \left. \left(-\cos(x) - \sin(x) \right) \right|_{\pi/4}^{3\pi/4} \\ &= \left[-\cos\left(\frac{3\pi}{4}\right) - \sin\left(\frac{3\pi}{4}\right) \right] - \left[-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) \right] \\ &= \left[ \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right] + \left[ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right] \\ &= \sqrt{2}\end{aligned}\].
3. We can rewrite \(y = 2x - 1\) as \(x = \frac{y+1}{2}\). The two curves intersect when \(y^2 - y - 2 = \frac{y+1}{2}\). It follows that \(2y^2 - 2y - 4 = y+1\), so \(2y^2 - 3y - 5 = 0\) and thus factoring, \((2y-5)(y+1) = 0\). Hence the two curves intersect when \(y = -1\) or \(y = \frac{5}{2}\). The figure below shows that \(x = \frac{y+1}{2}\) lies to the right of \(x = y^2 - y - 2\)on this interval, so the area of the region between the two curves is \[\begin{aligned}A &= \int_{-1}{5/2} \frac{y+1}{2} - (y^2-y-2) \ dy \\ &= \int_{-1}{5/2} -y^2 + \frac{3}{2}y + \frac{5}{2} \ dy \\ &= \left. \left( -\frac{y^3}{3} + \frac{3}{4}y^2 + \frac{5}{2}y \right) \right|_{-1}{5/2} \\ &= -\frac{1}{3}\left[\left(\frac{5}{2}\right)^3 + 1 \right] + \frac{3}{4}\left[\left(\frac{5}{2}\right)^2 - 1 \right] + \frac{5}{2}\left[\left(\frac{5}{2}\right) + 1 \right] \\ &= -\frac{133}{24} + \frac{63}{16} + \frac{35}{4} \\ &= \frac{343}{48}\end{aligned}\].
4. A representative picture (with \(m = 2\)) is shown in the figure below. The two curves intersect when \(x^2 - 1 = mx\), or when \(x^2 - mx - 1 = 0\). By the quadratic formula, it folows that \[\begin{aligned}\end{aligned}\]. If we integrate with vertical slices, then the graph of \(y = mx\) is always above the graph of \(y = x^2 - 1\) and so the area of the region between the two graphs is \[\begin{aligned}A &= \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx \\ &= \left. \left(-\frac{x^3}{3} + m\frac{x^2}{2} + x\right) \right|_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} \\ &= - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2 \\ &\qquad - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right)\end{aligned}\].
Let \(f(x) = 1-x^2\) and \(g(x) = ax^2 - a\), where \(a\) is an unknown positive real number. For what value(s) of \(a\) is the area between the curves \(f\) and \(g\) equal to 2?

Αποκάλυψέ την.

Both of these quadratic functions pass through the points \((-1,0)\) and \((1,0)\), and thus these are the two intersection points for all positive values of \(a\). In addition, since \(1 - x^2 \gt ax^2 - a\) on \(-1 \lt x \lt 1\), it follows that the area between the curves is given by \[\begin{aligned}\end{aligned}\]. Evaluating the integral, we find that \[\begin{aligned}A &= \int_{-1}^{1} \left( 1 + a - (1+a)x^2 \right) \, dx \\ &= \left. (1+a)x - \frac{1+a}{3}x^3 \right|_{-1}^{1} \\ &= (1+a)(1) - \frac{1+a}{3}(1)^3 - \left( (1+a)(-1) - \frac{1+a}{3}(-1)^3 \right) \\ &= 2(1+a) - \frac{2(1+a)}{3} \\ &= \frac{4(1+a)}{3}\end{aligned}\]. We want to know for which value(s) of \(a\) it follows that \(A = 2\). Thus, we solve \[\begin{aligned}\end{aligned}\], from which it follows that \(a = \frac{1}{2}\). Hence, in the figure below, the area between the two curves is exactly \(A = 2\).
Let \(f(x) = 2-x^2\). Recall that the average value of any continuous function \(f\) on an interval \([a,b]\) is given by \(\frac{1}{b-a} \int_a^b f(x) \, dx\).
1. Find the average value of \(f(x) = 2-x^2\) on the interval \([0,\sqrt{2}]\). Call this value \(r\).
2. Sketch a graph of \(y = f(x)\) and \(y = r\). Find their intersection point(s).
3. Show that on the interval \([0,\sqrt{2}]\), the amount of area that lies below \(y = f(x)\) and above \(y = r\) is equal to the amount of area that lies below \(y = r\) and above \(y = f(x)\).
4. Will the result of (c) be true for any continuous function and its average value on any interval? Why?
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1. The average value of \(f\) on the interval \([0,\sqrt{2}]\) is \[\begin{aligned}r &= \frac{1}{\sqrt{2}} \int_0^{\sqrt{2}} 2-x^2 \, dx \\ &= \left. \frac{1}{\sqrt{2}}\left(2x - \frac{x^3}{3}\right) \right|_0^{\sqrt{2}} \\ &= \frac{1}{\sqrt{2}}\left( 2\sqrt{2} - \frac{\sqrt{2}^3}{3} \right) \\ &= 2 - \frac{2}{3} \\ &= \frac{4}{3}\end{aligned}\].
2. The intersection point of \(y=f(x)\) and \(y=r\) occurs, for positive \(x\), when \(2-x^2 = \frac{4}{3}\), from which it follows that \(x^2 = \frac{2}{3}\) and thus \(x = \frac{\sqrt{6}}{3}\). A graph of \(f\), \(y = r\), and the intersection point is shown in the following figure.
3. Let \(A_1\) be the area that lies below \(y = f(x)\) and above \(y = r\) on the interval \(\left[0, \frac{\sqrt{6}}{3}\right]\) and \(A_2\) the are of the region that lies below \(y = r\) and above \(y = f(x)\) on \(\left[\frac{\sqrt{6}}{3}, \sqrt{2}\right]\) as illustrated in the figure below. Then \[\begin{aligned}A_1 &= \int_0^{\sqrt{6}/3} (2-x^2) - \frac{4}{3} \ dx \\ &= \int_0^{\sqrt{6}/3} \frac{2}{3} - x^2 \ dx \\ &= \left. \left( \frac{2}{3}x - \frac{x^3}{3} \right) \right|_0^{\sqrt{6}/3} \\ &= \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) - \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 \\ &= \frac{2\sqrt{6}}{9} - \frac{6 \sqrt{6}}{81} \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\]. and \[\begin{aligned}A_2 &= \int_{\sqrt{6}/3}^{\sqrt{2}} \frac{4}{3} - (2-x^2) \ dx \\ &= \int_{\sqrt{6}/3}^{\sqrt{2}} x^2 - \frac{2}{3} \ dx \\ &= \left. \left(\frac{x^3}{3} - \frac{2}{3}x \right) \right|_{\sqrt{6}/3}^{\sqrt{2}} \\ &= \left(\frac{2\sqrt{2}}{3} - \frac{2}{3} \sqrt{2} \right) - \left[ \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 - \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) \right] \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\]. Hence \(A_1 = A_2\) as desired.
4. The answer is yes. Let \(AV\) is the average value of a function \(f\) on an interval \([a,b]\). By the definition of average value, \[\begin{aligned}\end{aligned}\], which can also be viewed as the area of the rectangle with base \((b-a)\) and height \(AV\). Now suppose that the area of the region above the graph of \(f\) and below the line \(y=AV\) on the interval \([a,b]\) is \(R_1\) and the area of the region below the graph of \(f\) and above the line \(y=AV\) on the interval \([a,b]\) is \(R_2\). The figure below illustrates that the value of the definite integral is the area of the rectangle plus the area above the rectangle minus the area below the rectangle, or \[\begin{aligned}\end{aligned}\]. Having already established that \(AV(b-a) = \int_a^b f(x) \ dx\), it follows that \(R_2 - R_1 = 0\). So the area of the region that lies below \(y = f(x)\) and above \(y = AV\) is equal to the amount of area that lies below \(y = AV\) and above \(y = f(x)\) on \([a,b]\).

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Using definite integrals to find area and length

How can we use definite integrals to measure the area between two curves?
How do we decide whether to integrate with respect to x or with respect to y when we try to find the area of a region?
How can a definite integral be used to measure the length of a curve?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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