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Areas between Curves

Determine the area of a region between two curves by integrating with respect to the independent variable.

Area of a Region between Two Curves

Let \(f(x)\) and \(g(x)\) be continuous functions over an interval \([a,b]\) such that \(f(x)\ge g(x)\) on \([a,b].\) We want to find the area between the graphs of the functions, as shown in the following figure.

As we did before, we are going to partition the interval on the \(x\text{-axis}\) and approximate the area between the graphs of the functions with rectangles. So, for \(i=0,1,2\text{,\ldots },n,\) let \(P=\{{x}_{i}\}\) be a regular partition of \([a,b].\) Then, for \(i=1,2\text{,\ldots },n,\) choose a point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}],\) and on each interval \([{x}_{i-1},{x}_{i}]\) construct a rectangle that extends vertically from \(g({x}_{i}^{*})\) to \(f({x}_{i}^{*}).\) (a) shows the rectangles when \({x}_{i}^{*}\) is selected to be the left endpoint of the interval and \(n=10.\) (b) shows a representative rectangle in detail.

The height of each individual rectangle is \(f({x}_{i}^{*})-g({x}_{i}^{*})\) and the width of each rectangle is \(\text{\Delta }x.\) Adding the areas of all the rectangles, we see that the area between the curves is approximated by

\[A\approx \sum _{i=1}^{n}[f({x}_{i}^{*})-g({x}_{i}^{*})]\text{\Delta }x.\]

This is a Riemann sum, so we take the limit as \(n\to \infty\) and we get

\[A=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}[f({x}_{i}^{*})-g({x}_{i}^{*})]\text{\Delta }x={\int }_{a}^{b}[f(x)-g(x)]dx.\]

These findings are summarized in the following theorem.

We apply this theorem in the following example.

Example

Try it.

If R is the region bounded above by the graph of the function \(f(x)=x+4\) and below by the graph of the function \(g(x)=3-\frac{x}{2}\) over the interval \([1,4],\) find the area of region \(R.\)

Solution

The region is depicted in the following figure.

We have

\[\begin{array}{ll}A & ={\int }_{a}^{b}[f(x)-g(x)]dx \\ & ={\int }_{1}^{4}[(x+4)-(3-\frac{x}{2})]dx={\int }_{1}^{4}[\frac{3x}{2}+1]dx \\ & ={[\frac{3{x}^{2}}{4}+x]\ |}_{1}^{4}=(16-\frac{7}{4})=\frac{57}{4}.\end{array}\]

The area of the region is \(\frac{57}{4}\ {\text{units}}^{2}.\)

Condensed — the full section is in OpenStax Calculus Volume 2.

Areas of Compound Regions

So far, we have required \(f(x)\ge g(x)\) over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.

In practice, applying this theorem requires us to break up the interval \([a,b]\) and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.

Example

Try it.

If R is the region between the graphs of the functions \(f(x)=\text{sin}\ x\) and \(g(x)=\text{cos}\ x\) over the interval \([0,\pi ],\) find the area of region \(R.\)

Solution

The region is depicted in the following figure.

The graphs of the functions intersect at \(x=\pi \text{/}4.\) For \(x\in [0,\pi \text{/}4],\) \(\text{cos}\ x\ge \text{sin}\ x,\) so

\[|f(x)-g(x)|=|\text{sin}\ x-\text{cos}\ x|=\text{cos}\ x-\text{sin}\ x.\]

On the other hand, for \(x\in [\pi \text{/}4,\pi ],\) \(\text{sin}\ x\ge \text{cos}\ x,\) so

\[|f(x)-g(x)|=|\text{sin}\ x-\text{cos}\ x|=\text{sin}\ x-\text{cos}\ x.\]

Then

\[\begin{array}{ll}A & ={\int }_{a}^{b}|f(x)-g(x)|dx \\ & ={\int }_{0}^{\pi }|\text{sin}\ x-\text{cos}\ x|dx={\int }_{0}^{\pi \text{/}4}(\text{cos}\ x-\text{sin}\ x)dx+{\int }_{\pi \text{/}4}^{\pi }(\text{sin}\ x-\text{cos}\ x)dx \\ & ={[\text{sin}\ x+\text{cos}\ x]\ |}_{0}^{\pi \text{/}4}+{[\text{-}\text{cos}\ x-\text{sin}\ x]\ |}_{\pi \text{/}4}^{\pi } \\ & =(\sqrt{2}-1)+(1+\sqrt{2})=2\sqrt{2}.\end{array}\]

The area of the region is \(2\sqrt{2}\) units².

Condensed — the full section is in OpenStax Calculus Volume 2.

Regions Defined with Respect to

In , we had to evaluate two separate integrals to calculate the area of the region. However, there is another approach that requires only one integral. What if we treat the curves as functions of \(y,\) instead of as functions of \(x?\) Review . Note that the left graph, shown in red, is represented by the function \(y=f(x)={x}^{2}.\) We could just as easily solve this for \(x\) and represent the curve by the function \(x=v(y)=\sqrt{y}.\) (Note that \(x=\text{-}\sqrt{y}\) is also a valid representation of the function \(y=f(x)={x}^{2}\) as a function of \(y.\) However, based on the graph, it is clear we are interested in the positive square root.) Similarly, the right graph is represented by the function \(y=g(x)=2-x,\) but could just as easily be represented by the function \(x=u(y)=2-y.\) When the graphs are represented as functions of \(y,\) we see the region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, if we integrate with respect to \(y,\) we need to evaluate one integral only. Let’s develop a formula for this type of integration.

Let \(u(y)\) and \(v(y)\) be continuous functions over an interval \([c,d]\) such that \(u(y)\ge v(y)\) for all \(y\in [c,d].\) We want to find the area between the graphs of the functions, as shown in the following figure.

This time, we are going to partition the interval on the \(y\text{-axis}\) and use horizontal rectangles to approximate the area between the functions. So, for \(i=0,1,2\text{,\ldots },n,\) let \(Q=\{{y}_{i}\}\) be a regular partition of \([c,d].\) Then, for \(i=1,2\text{,\ldots },n,\) choose a point \({y}_{i}^{*}\in [{y}_{i-1},{y}_{i}],\) then over each interval \([{y}_{i-1},{y}_{i}]\) construct a rectangle that extends horizontally from \(v({y}_{i}^{*})\) to \(u({y}_{i}^{*}).\) (a) shows the rectangles when \({y}_{i}^{*}\) is selected to be the lower endpoint of the interval and \(n=10.\) (b) shows a representative rectangle in detail.

The height of each individual rectangle is \(\text{\Delta }y\) and the width of each rectangle is \(u({y}_{i}^{*})-v({y}_{i}^{*}).\) Therefore, the area between the curves is approximately

\[A\approx \sum _{i=1}^{n}[u({y}_{i}^{*})-v({y}_{i}^{*})]\text{\Delta }y.\]

This is a Riemann sum, so we take the limit as \(n\to \infty ,\) obtaining

\[A=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}[u({y}_{i}^{*})-v({y}_{i}^{*})]\text{\Delta }y={\int }_{c}^{d}[u(y)-v(y)]dy.\]

These findings are summarized in the following theorem.

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

Just as definite integrals can be used to find the area under a curve, they can also be used to find the area between two curves.
To find the area between two curves defined by functions, integrate the difference of the functions.
If the graphs of the functions cross, or if the region is complex, use the absolute value of the difference of the functions. In this case, it may be necessary to evaluate two or more integrals and add the results to find the area of the region.
Sometimes it can be easier to integrate with respect to y to find the area. The principles are the same regardless of which variable is used as the variable of integration.

Key Equations

Area between two curves, integrating on the x-axis	\(A={\int }_{a}^{b}[f(x)-g(x)]dx\)
Area between two curves, integrating on the y-axis	\(A={\int }_{c}^{d}[u(y)-v(y)]dy\)

Areas between Curves

For the following exercises, determine the area of the region between the two curves in the given figure by integrating over the \(x\text{-axis}\text{.}\)

For the following exercises, split the region between the two curves into two smaller regions, then determine the area by integrating over the \(x\text{-axis}.\) Note that you will have two integrals to solve.

For the following exercises, determine the area of the region between the two curves by integrating over the \(y\text{-axis}.\)

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(x\text{-axis}.\)

For the following exercises, graph the equations and shade the area of the region between the curves. If necessary, break the region into sub-regions to determine its entire area.

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(y\text{-axis}.\)

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the x-axis or y-axis, whichever seems more convenient.

Condensed — the full section is in OpenStax Calculus Volume 2.

Area of a Region between Two Curves

\[A\approx \sum _{i=1}^{n}[f({x}_{i}^{*})-g({x}_{i}^{*})]\text{\Delta }x.\]

This is a Riemann sum, so we take the limit as \(n\to \infty\) and we get

\[A=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}[f({x}_{i}^{*})-g({x}_{i}^{*})]\text{\Delta }x={\int }_{a}^{b}[f(x)-g(x)]dx.\]

These findings are summarized in the following theorem.

We apply this theorem in the following example.

Example

Try it.

If R is the region bounded above by the graph of the function \(f(x)=x+4\) and below by the graph of the function \(g(x)=3-\frac{x}{2}\) over the interval \([1,4],\) find the area of region \(R.\)

Solution

The region is depicted in the following figure.

We have

The area of the region is \(\frac{57}{4}\ {\text{units}}^{2}.\)

Condensed — the full section is in OpenStax Calculus Volume 1.

Areas of Compound Regions

Example

Try it.

If R is the region between the graphs of the functions \(f(x)=\text{sin}\ x\) and \(g(x)=\text{cos}\ x\) over the interval \([0,\pi ],\) find the area of region \(R.\)

Solution

The region is depicted in the following figure.

The graphs of the functions intersect at \(x=\pi \text{/}4.\) For \(x\in [0,\pi \text{/}4],\) \(\text{cos}\ x\ge \text{sin}\ x,\) so

\[|f(x)-g(x)|=|\text{sin}\ x-\text{cos}\ x|=\text{cos}\ x-\text{sin}\ x.\]

On the other hand, for \(x\in [\pi \text{/}4,\pi ],\) \(\text{sin}\ x\ge \text{cos}\ x,\) so

\[|f(x)-g(x)|=|\text{sin}\ x-\text{cos}\ x|=\text{sin}\ x-\text{cos}\ x.\]

Then

The area of the region is \(2\sqrt{2}\) units².

Condensed — the full section is in OpenStax Calculus Volume 1.

Regions Defined with Respect to

The height of each individual rectangle is \(\text{\Delta }y\) and the width of each rectangle is \(u({y}_{i}^{*})-v({y}_{i}^{*}).\) Therefore, the area between the curves is approximately

\[A\approx \sum _{i=1}^{n}[u({y}_{i}^{*})-v({y}_{i}^{*})]\text{\Delta }y.\]

This is a Riemann sum, so we take the limit as \(n\to \infty ,\) obtaining

\[A=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}[u({y}_{i}^{*})-v({y}_{i}^{*})]\text{\Delta }y={\int }_{c}^{d}[u(y)-v(y)]dy.\]

These findings are summarized in the following theorem.

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

Just as definite integrals can be used to find the area under a curve, they can also be used to find the area between two curves.
To find the area between two curves defined by functions, integrate the difference of the functions.
If the graphs of the functions cross, or if the region is complex, use the absolute value of the difference of the functions. In this case, it may be necessary to evaluate two or more integrals and add the results to find the area of the region.
Sometimes it can be easier to integrate with respect to y to find the area. The principles are the same regardless of which variable is used as the variable of integration.

Key Equations

Area between two curves, integrating on the x-axis	\(A={\int }_{a}^{b}[f(x)-g(x)]dx\)
Area between two curves, integrating on the y-axis	\(A={\int }_{c}^{d}[u(y)-v(y)]dy\)

Areas between Curves

For the following exercises, determine the area of the region between the two curves in the given figure by integrating over the \(x\text{-axis}\text{.}\)

For the following exercises, determine the area of the region between the two curves by integrating over the \(y\text{-axis}.\)

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(x\text{-axis}.\)

For the following exercises, graph the equations and shade the area of the region between the curves. If necessary, break the region into sub-regions to determine its entire area.

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(y\text{-axis}.\)

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the x-axis or y-axis, whichever seems more convenient.

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

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

If R is the region bounded above by the graph of the function \(f(x)=x+4\) and below by the graph of the function \(g(x)=3-\frac{x}{2}\) over the interval \([1,4],\) find the area of region \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

The region is depicted in the following figure.
We have
\[\begin{array}{ll}A & ={\int }_{a}^{b}[f(x)-g(x)]dx \\ & ={\int }_{1}^{4}[(x+4)-(3-\frac{x}{2})]dx={\int }_{1}^{4}[\frac{3x}{2}+1]dx \\ & ={[\frac{3{x}^{2}}{4}+x]\ |}_{1}^{4}=(16-\frac{7}{4})=\frac{57}{4}.\end{array}\]
The area of the region is \(\frac{57}{4}\ {\text{units}}^{2}.\)
If \(R\) is the region bounded by the graphs of the functions \(f(x)=\frac{x}{2}+5\) and \(g(x)=x+\frac{1}{2}\) over the interval \([1,5],\) find the area of region \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(12\) units²
If \(R\) is the region bounded above by the graph of the function \(f(x)=9-{(x\text{/}2)}^{2}\) and below by the graph of the function \(g(x)=6-x,\) find the area of region \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

The region is depicted in the following figure.
We first need to compute where the graphs of the functions intersect. Setting \(f(x)=g(x),\) we get
\[\begin{array}{lll}f(x) & = & g(x) \\ \\ 9-{(\frac{x}{2})}^{2} & = & 6-x \\ 9-\frac{{x}^{2}}{4} & = & 6-x \\ 36-{x}^{2} & = & 24-4x \\ {x}^{2}-4x-12 & = & 0 \\ (x-6)(x+2) & = & 0.\end{array}\]
The graphs of the functions intersect when \(x=6\) or \(x=-2,\) so we want to integrate from \(-2\) to \(6.\) Since \(f(x)\ge g(x)\) for \(-2\le x\le 6,\) we obtain
\[\begin{array}{ll}A & ={\int }_{a}^{b}[f(x)-g(x)]dx \\ & ={\int }_{-2}^{6}[9-{(\frac{x}{2})}^{2}-(6-x)]dx={\int }_{-2}^{6}[3-\frac{{x}^{2}}{4}+x]dx \\ & ={[3x-\frac{{x}^{3}}{12}+\frac{{x}^{2}}{2}]\ |}_{-2}^{6}=\frac{64}{3}.\end{array}\]
The area of the region is \(64\text{/}3\) units².
If R is the region bounded above by the graph of the function \(f(x)=x\) and below by the graph of the function \(g(x)={x}^{4},\) find the area of region \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{3}{10}\) unit²
If R is the region between the graphs of the functions \(f(x)=\text{sin}\ x\) and \(g(x)=\text{cos}\ x\) over the interval \([0,\pi ],\) find the area of region \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

The region is depicted in the following figure.
The graphs of the functions intersect at \(x=\pi \text{/}4.\) For \(x\in [0,\pi \text{/}4],\) \(\text{cos}\ x\ge \text{sin}\ x,\) so
\[|f(x)-g(x)|=|\text{sin}\ x-\text{cos}\ x|=\text{cos}\ x-\text{sin}\ x.\]
On the other hand, for \(x\in [\pi \text{/}4,\pi ],\) \(\text{sin}\ x\ge \text{cos}\ x,\) so
\[|f(x)-g(x)|=|\text{sin}\ x-\text{cos}\ x|=\text{sin}\ x-\text{cos}\ x.\]
Then
\[\begin{array}{ll}A & ={\int }_{a}^{b}|f(x)-g(x)|dx \\ & ={\int }_{0}^{\pi }|\text{sin}\ x-\text{cos}\ x|dx={\int }_{0}^{\pi \text{/}4}(\text{cos}\ x-\text{sin}\ x)dx+{\int }_{\pi \text{/}4}^{\pi }(\text{sin}\ x-\text{cos}\ x)dx \\ & ={[\text{sin}\ x+\text{cos}\ x]\ |}_{0}^{\pi \text{/}4}+{[\text{-}\text{cos}\ x-\text{sin}\ x]\ |}_{\pi \text{/}4}^{\pi } \\ & =(\sqrt{2}-1)+(1+\sqrt{2})=2\sqrt{2}.\end{array}\]
The area of the region is \(2\sqrt{2}\) units².
If R is the region between the graphs of the functions \(f(x)=\text{sin}\ x\) and \(g(x)=\text{cos}\ x\) over the interval \([\pi \text{/}2,2\pi ],\) find the area of region \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(2+2\sqrt{2}\) units²
Consider the region depicted in . Find the area of \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

As with , we need to divide the interval into two pieces. The graphs of the functions intersect at \(x=1\) (set \(f(x)=g(x)\) and solve for x), so we evaluate two separate integrals: one over the interval \([0,1]\) and one over the interval \([1,2].\)
Over the interval \([0,1],\) the region is bounded above by \(f(x)={x}^{2}\) and below by the x-axis, so we have
\[{A}_{1}={\int }_{0}^{1}{x}^{2}dx={\frac{{x}^{3}}{3}\ |}_{0}^{1}=\frac{1}{3}.\]
Over the interval \([1,2],\) the region is bounded above by \(g(x)=2-x\) and below by the \(x\text{-axis,}\) so we have
\[{A}_{2}={\int }_{1}^{2}(2-x)dx={[2x-\frac{{x}^{2}}{2}]\ |}_{1}^{2}=\frac{1}{2}.\]
Adding these areas together, we obtain
\[A={A}_{1}+{A}_{2}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}.\]
The area of the region is \(5\text{/}6\) units².
Consider the region depicted in the following figure. Find the area of \(R.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{5}{3}\) units²
Let’s revisit , only this time let’s integrate with respect to \(y.\) Let \(R\) be the region depicted in . Find the area of \(R\) by integrating with respect to \(y.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

We must first express the graphs as functions of \(y.\) As we saw at the beginning of this section, the curve on the left can be represented by the function \(x=v(y)=\sqrt{y},\) and the curve on the right can be represented by the function \(x=u(y)=2-y.\)
Now we have to determine the limits of integration. The region is bounded below by the x-axis, so the lower limit of integration is \(y=0.\) The upper limit of integration is determined by the point where the two graphs intersect, which is the point \((1,1),\) so the upper limit of integration is \(y=1.\) Thus, we have \([c,d]=[0,1].\)
Calculating the area of the region, we get
\[\begin{array}{ll}A & ={\int }_{c}^{d}[u(y)-v(y)]dy \\ & ={\int }_{0}^{1}[(2-y)-\sqrt{y}]dy={[2y-\frac{{y}^{2}}{2}-\frac{2}{3}{y}^{3\text{/}2}]\ |}_{0}^{1} \\ & =\frac{5}{6}.\end{array}\]
The area of the region is \(5\text{/}6\) units².
Let’s revisit the checkpoint associated with , only this time, let’s integrate with respect to \(y.\) Let \(R\) be the region depicted in the following figure. Find the area of \(R\) by integrating with respect to \(y.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{5}{3}\) units²
\(y={x}^{2}-3\ \text{and}\ y=1\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{32}{3}\)
\(y={x}^{2}\ \text{and}\ y=3x+4\)
\(y={x}^{3}\) and \(y={x}^{2}+x\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{13}{12}\)
\(y=\text{cos}\ \theta\) and \(y=0.5,\) for \(0\le \theta \le \pi\)
\(x={y}^{2}\ \text{and}\ x=9\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(36\)
\(y=x\ \text{and}\ x={y}^{2}\)
\(y={x}^{2}\ \text{and}\ y=\text{-}{x}^{2}+18x\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

243 square units
\(y=\frac{1}{x},y=\frac{1}{{x}^{2}},\ \text{and}\ x=3\)
\(y=\text{cos}\ x\) and \(y={\text{cos}}^{2}x\) on \(x=[\text{-}\pi ,\pi ]\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

4
\(y={e}^{x},y={e}^{2x-1},\ \text{and}\ x=0\)
\(y={e}^{x},y={e}^{\text{-}x},x=-1\ \text{and}\ x=1\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{2{(e-1)}^{2}}{e}\)
\(y=e,y={e}^{x},\ \text{and}\ y={e}^{\text{-}x}\)
\(y=|x|\ \text{and}\ y={x}^{2}\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{1}{3}\)
\(y=\text{sin}(\pi x),y=2x,\ \text{and}\ x>0\)
\(y=12-x,y=\sqrt{x},\ \text{and}\ y=1\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{34}{3}\)
\(y=\text{sin}\ x\) and \(y=\text{cos}\ x\) over \(x=[\text{-}\pi ,\pi ]\)
\(y={x}^{3}\ \text{and}\ y={x}^{2}-2x\) over \(x=[-1,1]\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{5}{2}\)
\(y={x}^{2}+9\ \text{and}\ y=10+2x\) over \(x=[-1,3]\)
\(y={x}^{3}+3x\) and \(y=4x\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{1}{2}\)
\(x={y}^{3}\ \text{and}\ x=3y-2\)
\(x=2y\ \text{and}\ x={y}^{3}-y\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{9}{2}\)
\(x=-3+{y}^{2}\ \text{and}\ x=y-{y}^{2}\)
\({y}^{2}=x\ \text{and}\ x=y+2\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{9}{2}\)
\(x=|y|\ \text{and}\ 2x=\text{-}{y}^{2}+2\)
\(x=\text{sin}\ y,x=\text{cos}(2y),y=\pi \text{/}2,\text{and}\ y=\text{-}\pi \text{/}2\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{3\sqrt{3}}{2}\)
\(x={y}^{4}\text{and}\ x={y}^{5}\)
\(y=x{e}^{x},y={e}^{x},x=0,\ \text{and}\ x=1\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(e-2\)
\(y={x}^{6}\text{and}\ y={x}^{4}\)
\(x={y}^{3}+2{y}^{2}+1\ \text{and}\ x=\text{-}{y}^{2}+1\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{27}{4}\)
\(y=|x|\ \text{and}\ y={x}^{2}-1\)

Symbols used here

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

summation
Add a_k for k = 1 up to n.

\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…

\theta

theta
The usual name for an angle.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

\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.

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

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

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

derivative
Instantaneous rate of change; slope of the graph.

How to: Areas between Curves

Determine the area of a region between two curves by integrating with respect to the independent variable.
Find the area of a compound region.
Determine the area of a region between two curves by integrating with respect to the dependent variable.
Just as definite integrals can be used to find the area under a curve, they can also be used to find the area between two curves.
To find the area between two curves defined by functions, integrate the difference of the functions.
If the graphs of the functions cross, or if the region is complex, use the absolute value of the difference of the functions. In this case, it may be necessary to evaluate two or more integrals and add the results to find the area of the region.
Sometimes it can be easier to integrate with respect to

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 OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ଅଧିକ Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests