maths.free › Calculus › Applications of Integration › Integrals, Exponential Functions, and Logarithms

Integrals, Exponential Functions, and Logarithms

Write the definition of the natural logarithm as an integral.

The Natural Logarithm as an Integral

Recall the power rule for integrals:

\[\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C,\ n\ne \text{-}1.\]

Clearly, this does not work when \(n=-1,\) as it would force us to divide by zero. So, what do we do with \(\int \frac{1}{x}dx?\) Recall from the Fundamental Theorem of Calculus that \({\int }_{1}^{x}\frac{1}{t}dt\) is an antiderivative of \(1\text{/}x.\) Therefore, we can make the following definition.

For \(x>1,\) this is just the area under the curve \(y=1\text{/}t\) from \(1\) to \(x.\) For \(x<1,\) we have \({\int }_{1}^{x}\frac{1}{t}dt=\text{-}{\int }_{x}^{1}\frac{1}{t}dt,\) so in this case it is the negative of the area under the curve from \(x\ \text{to}\ 1\) (see the following figure).

Notice that \(\text{ln}\ 1=0.\) Furthermore, the function \(y=1\text{/}t>0\) for \(x>0.\) Therefore, by the properties of integrals, it is clear that \(\text{ln}\ x\) is increasing for \(x>0.\)

Properties of the Natural Logarithm

Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.

A graph of \(\text{ln}\ x\) is Notice that it is continuous throughout its domain of \((0,\infty ).\)

Example

Try it.

Calculate the following derivatives:

\(\frac{d}{dx}\text{ln}(5{x}^{3}-2)\)
\(\frac{d}{dx}{(\text{ln}(3x))}^{2}\)

Solution

We need to apply the chain rule in both cases.

\(\frac{d}{dx}\text{ln}(5{x}^{3}-2)=\frac{15{x}^{2}}{5{x}^{3}-2}\)
\(\frac{d}{dx}{(\text{ln}(3x))}^{2}=\frac{2(\text{ln}(3x))\cdot 3}{3x}=\frac{2(\text{ln}(3x))}{x}\)

Note that if we use the absolute value function and create a new function \(\text{ln}\ |x|,\) we can extend the domain of the natural logarithm to include \(x<0.\) Then \((d\text{/}(dx))\text{ln}\ |x|=1\text{/}x.\) This gives rise to the familiar integration formula.

Example

Try it.

Calculate the integral \(\int \frac{x}{{x}^{2}+4}dx.\)

Solution

Using \(u\)-substitution, let \(u={x}^{2}+4.\) Then \(du=2x\ dx\) and we have

\[\int \frac{x}{{x}^{2}+4}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\text{ln}\ |u|+C=\frac{1}{2}\text{ln}\ |{x}^{2}+4|+C=\frac{1}{2}\text{ln}({x}^{2}+4)+C.\]

Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

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

Defining the Number

Now that we have the natural logarithm defined, we can use that function to define the number \(e.\)

To put it another way, the area under the curve \(y=1\text{/}t\) between \(t=1\) and \(t=e\) is \(1\) (). The proof that such a number exists and is unique is left to you. (Hint: Use the Intermediate Value Theorem to prove existence and the fact that \(\text{ln}\ x\) is increasing to prove uniqueness.)

The number \(e\) can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series). Its approximate value is given by

\[e\approx 2.71828182846.\]

The Exponential Function

We now turn our attention to the function \({e}^{x}.\) Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by \(\text{exp}\ x.\) Then,

\[\text{exp}(\text{ln}\ x)=x\ \text{for}\ x>0\ \text{and}\ \text{ln}(\text{exp}\ x)=x\ \text{for all}\ x.\]

The following figure shows the graphs of \(\text{exp}\ x\) and \(\text{ln}\ x.\)

We hypothesize that \(\text{exp}\ x={e}^{x}.\) For rational values of \(x,\) this is easy to show. If \(x\) is rational, then we have \(\text{ln}({e}^{x})=x\ \text{ln}\ e=x.\) Thus, when \(x\) is rational, \({e}^{x}=\text{exp}\ x.\) For irrational values of \(x,\) we simply define \({e}^{x}\) as the inverse function of \(\text{ln}\ x.\)

Then we have \({e}^{x}=\text{exp}(x)\) for all \(x,\) and thus

\[{e}^{\text{ln}\ x}=x\ \text{for}\ x>0\ \text{and}\ \text{ln}({e}^{x})=x\]

for all \(x.\)

Properties of the Exponential Function

Since the exponential function was defined in terms of an inverse function, and not in terms of a power of \(e,\) we must verify that the usual laws of exponents hold for the function \({e}^{x}.\)

Note that if \(p\) and \(q\) are rational, the properties hold. However, if \(p\) or \(q\) are irrational, we must apply the inverse function definition of \({e}^{x}\) and verify the properties. Only the first property is verified here; the other two are left to you. We have

\[\text{ln}({e}^{p}{e}^{q})=\text{ln}({e}^{p})+\text{ln}({e}^{q})=p+q=\text{ln}({e}^{p+q}).\]

Since \(\text{ln}\ x\) is one-to-one, then

\[{e}^{p}{e}^{q}={e}^{p+q}.\]

□

As with part iv. of the logarithm properties, we can extend property iii. to irrational values of \(r,\) and we do so by the end of the section.

We also want to verify the differentiation formula for the function \(y={e}^{x}.\) To do this, we need to use implicit differentiation. Let \(y={e}^{x}.\) Then

\[\begin{array}{lll}\text{ln}\ y & = & x \\ \frac{d}{dx}\text{ln}\ y & = & \frac{d}{dx}x \\ \frac{1}{y}\ \frac{dy}{dx} & = & 1 \\ \frac{dy}{dx} & = & y.\end{array}\]

Thus, we

\[\frac{d}{dx}{e}^{x}={e}^{x}\]

as desired, which leads immediately to the integration formula

\[\int {e}^{x}dx={e}^{x}+C.\]

We apply these formulas in the following examples.

Example

Try it.

Evaluate the following derivatives:

\(\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}\)
\(\frac{d}{dx}{e}^{3{x}^{2}}\)

Solution

We apply the chain rule as necessary.

\(\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}=\frac{d}{dt}{e}^{3t+{t}^{2}}={e}^{3t+{t}^{2}}(3+2t)\)
\(\frac{d}{dx}{e}^{3{x}^{2}}={e}^{3{x}^{2}}6x\)

Example

Try it.

Evaluate the following integral: \(\int 2x{e}^{\text{-}{x}^{2}}dx.\)

Solution

Using \(u\)-substitution, let \(u=\text{-}{x}^{2}.\) Then \(du=-2x\ dx,\) and we have

\[\int 2x{e}^{\text{-}{x}^{2}}dx=\text{-}\int {e}^{u}du=\text{-}{e}^{u}+C=\text{-}{e}^{\text{-}{x}^{2}}+C.\]

General Logarithmic and Exponential Functions

We close this section by looking at exponential functions and logarithms with bases other than \(e.\) Exponential functions are functions of the form \(f(x)={a}^{x}.\) Note that unless \(a=e,\) we still do not have a mathematically rigorous definition of these functions for irrational exponents. Let’s rectify that here by defining the function \(f(x)={a}^{x}\) in terms of the exponential function \({e}^{x}.\) We then examine logarithms with bases other than \(e\) as inverse functions of exponential functions.

Now \({a}^{x}\) is defined rigorously for all values of x. This definition also allows us to generalize property iv. of logarithms and property iii. of exponential functions to apply to both rational and irrational values of \(r.\) It is straightforward to show that properties of exponents hold for general exponential functions defined in this way.

Let’s now apply this definition to calculate a differentiation formula for \({a}^{x}.\) We have

\[\frac{d}{dx}{a}^{x}=\frac{d}{dx}{e}^{x\ \text{ln}\ a}={e}^{x\ \text{ln}\ a}\text{ln}\ a={a}^{x}\text{ln}\ a.\]

The corresponding integration formula follows immediately.

If \(a\ne 1,\) then the function \({a}^{x}\) is one-to-one and has a well-defined inverse. Its inverse is denoted by \({\text{log}}_{a}x.\) Then,

\[y={\text{log}}_{a}x\ \text{if and only if}\ x={a}^{y}.\]

Note that general logarithm functions can be written in terms of the natural logarithm. Let \(y={\text{log}}_{a}x.\) Then, \(x={a}^{y}.\) Taking the natural logarithm of both sides of this second equation, we get

\[\begin{array}{lll}\text{ln}\ x & = & \text{ln}({a}^{y}) \\ \text{ln}\ x & = & y\ \text{ln}\ a \\ y & = & \frac{\text{ln}\ x}{\text{ln}\ a} \\ {\text{log}}_{a}x & = & \frac{\text{ln}\ x}{\text{ln}\ a}.\end{array}\]

Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base \(a.\) Again, let \(y={\text{log}}_{a}x.\) Then,

\[\begin{array}{ll}\frac{dy}{dx} & =\frac{d}{dx}({\text{log}}_{a}x) \\ & =\frac{d}{dx}(\frac{\text{ln}\ x}{\text{ln}\ a}) \\ & =(\frac{1}{\text{ln}\ a})\frac{d}{dx}(\text{ln}\ x) \\ & =\frac{1}{\text{ln}\ a}\cdot \frac{1}{x} \\ & =\frac{1}{x\ \text{ln}\ a}.\end{array}\]

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

Key Concepts

The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally. This section develops the concepts in a mathematically rigorous way.
The cornerstone of the development is the definition of the natural logarithm in terms of an integral.
The function \({e}^{x}\) is then defined as the inverse of the natural logarithm.
General exponential functions are defined in terms of \({e}^{x},\) and the corresponding inverse functions are general logarithms.
Familiar properties of logarithms and exponents still hold in this more rigorous context.

Key Equations

Natural logarithm function	\(\text{ln}\ x={\int }_{1}^{x}\frac{1}{t}dt\) Z
Exponential function \(y={e}^{x}\)	\(\text{ln}\ y=\text{ln}({e}^{x})=x\) Z

Integrals, Exponential Functions, and Logarithms

For the following exercises, find the derivative \(\frac{dy}{dx}.\)

For the following exercises, find the indefinite integral.

For the following exercises, find the derivative \(dy\text{/}dx.\) (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

For the following exercises, find the definite or indefinite integral.

For the following exercises, compute \(dy\text{/}dx\) by differentiating \(\text{ln}\ y.\)

For the following exercises, evaluate by any method.

For the following exercises, use the function \(\text{ln}\ x.\) If you are unable to find intersection points analytically, use a calculator.

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

The Natural Logarithm as an Integral

Recall the power rule for integrals:

\[\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C,\ n\ne \text{-}1.\]

Notice that \(\text{ln}\ 1=0.\) Furthermore, the function \(y=1\text{/}t>0\) for \(x>0.\) Therefore, by the properties of integrals, it is clear that \(\text{ln}\ x\) is increasing for \(x>0.\)

Properties of the Natural Logarithm

Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.

A graph of \(\text{ln}\ x\) is Notice that it is continuous throughout its domain of \((0,\infty ).\)

Example

Try it.

Calculate the following derivatives:

\(\frac{d}{dx}\text{ln}(5{x}^{3}-2)\)
\(\frac{d}{dx}{(\text{ln}(3x))}^{2}\)

Solution

We need to apply the chain rule in both cases.

\(\frac{d}{dx}\text{ln}(5{x}^{3}-2)=\frac{15{x}^{2}}{5{x}^{3}-2}\)
\(\frac{d}{dx}{(\text{ln}(3x))}^{2}=\frac{2(\text{ln}(3x))\cdot 3}{3x}=\frac{2(\text{ln}(3x))}{x}\)

Example

Try it.

Calculate the integral \(\int \frac{x}{{x}^{2}+4}dx.\)

Solution

Using \(u\)-substitution, let \(u={x}^{2}+4.\) Then \(du=2x\ dx\) and we have

\[\int \frac{x}{{x}^{2}+4}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\text{ln}\ |u|+C=\frac{1}{2}\text{ln}\ |{x}^{2}+4|+C=\frac{1}{2}\text{ln}({x}^{2}+4)+C.\]

Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

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

Defining the Number

Now that we have the natural logarithm defined, we can use that function to define the number \(e.\)

The number \(e\) can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series). Its approximate value is given by

\[e\approx 2.71828182846.\]

The Exponential Function

\[\text{exp}(\text{ln}\ x)=x\ \text{for}\ x>0\ \text{and}\ \text{ln}(\text{exp}\ x)=x\ \text{for all}\ x.\]

The following figure shows the graphs of \(\text{exp}\ x\) and \(\text{ln}\ x.\)

Then we have \({e}^{x}=\text{exp}(x)\) for all \(x,\) and thus

\[{e}^{\text{ln}\ x}=x\ \text{for}\ x>0\ \text{and}\ \text{ln}({e}^{x})=x\]

for all \(x.\)

Properties of the Exponential Function

Since the exponential function was defined in terms of an inverse function, and not in terms of a power of \(e,\) we must verify that the usual laws of exponents hold for the function \({e}^{x}.\)

\[\text{ln}({e}^{p}{e}^{q})=\text{ln}({e}^{p})+\text{ln}({e}^{q})=p+q=\text{ln}({e}^{p+q}).\]

Since \(\text{ln}\ x\) is one-to-one, then

\[{e}^{p}{e}^{q}={e}^{p+q}.\]

□

As with part iv. of the logarithm properties, we can extend property iii. to irrational values of \(r,\) and we do so by the end of the section.

We also want to verify the differentiation formula for the function \(y={e}^{x}.\) To do this, we need to use implicit differentiation. Let \(y={e}^{x}.\) Then

\[\begin{array}{lll}\text{ln}\ y & = & x \\ \frac{d}{dx}\text{ln}\ y & = & \frac{d}{dx}x \\ \frac{1}{y}\ \frac{dy}{dx} & = & 1 \\ \frac{dy}{dx} & = & y.\end{array}\]

Thus, we

\[\frac{d}{dx}{e}^{x}={e}^{x}\]

as desired, which leads immediately to the integration formula

\[\int {e}^{x}dx={e}^{x}+C.\]

We apply these formulas in the following examples.

Example

Try it.

Evaluate the following derivatives:

\(\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}\)
\(\frac{d}{dx}{e}^{3{x}^{2}}\)

Solution

We apply the chain rule as necessary.

\(\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}=\frac{d}{dt}{e}^{3t+{t}^{2}}={e}^{3t+{t}^{2}}(3+2t)\)
\(\frac{d}{dx}{e}^{3{x}^{2}}={e}^{3{x}^{2}}6x\)

Example

Try it.

Evaluate the following integral: \(\int 2x{e}^{\text{-}{x}^{2}}dx.\)

Solution

Using \(u\)-substitution, let \(u=\text{-}{x}^{2}.\) Then \(du=-2x\ dx,\) and we have

\[\int 2x{e}^{\text{-}{x}^{2}}dx=\text{-}\int {e}^{u}du=\text{-}{e}^{u}+C=\text{-}{e}^{\text{-}{x}^{2}}+C.\]

General Logarithmic and Exponential Functions

Let’s now apply this definition to calculate a differentiation formula for \({a}^{x}.\) We have

\[\frac{d}{dx}{a}^{x}=\frac{d}{dx}{e}^{x\ \text{ln}\ a}={e}^{x\ \text{ln}\ a}\text{ln}\ a={a}^{x}\text{ln}\ a.\]

The corresponding integration formula follows immediately.

If \(a\ne 1,\) then the function \({a}^{x}\) is one-to-one and has a well-defined inverse. Its inverse is denoted by \({\text{log}}_{a}x.\) Then,

\[y={\text{log}}_{a}x\ \text{if and only if}\ x={a}^{y}.\]

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

Key Concepts

The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally. This section develops the concepts in a mathematically rigorous way.
The cornerstone of the development is the definition of the natural logarithm in terms of an integral.
The function \({e}^{x}\) is then defined as the inverse of the natural logarithm.
General exponential functions are defined in terms of \({e}^{x},\) and the corresponding inverse functions are general logarithms.
Familiar properties of logarithms and exponents still hold in this more rigorous context.

Key Equations

Natural logarithm function	\(\text{ln}\ x={\int }_{1}^{x}\frac{1}{t}dt\) Z
Exponential function \(y={e}^{x}\)	\(\text{ln}\ y=\text{ln}({e}^{x})=x\) Z

Integrals, Exponential Functions, and Logarithms

For the following exercises, find the derivative \(\frac{dy}{dx}.\)

For the following exercises, find the indefinite integral.

For the following exercises, find the derivative \(dy\text{/}dx.\) (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

For the following exercises, find the definite or indefinite integral.

For the following exercises, compute \(dy\text{/}dx\) by differentiating \(\text{ln}\ y.\)

For the following exercises, evaluate by any method.

For the following exercises, use the function \(\text{ln}\ x.\) If you are unable to find intersection points analytically, use a calculator.

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

Haddaad Ma jiro xisaabis oo ku nool midkan, laakiin qaybaha ka mid ah waa la xisaabi karaa. Ku raaxayso mid ka hooseeya, ama ku qor adiga kuu gaar ah.

Tababar (40)

Ku day mid kasta oo ka mid ah oo ku saabsan buugaag hore. Muuji jawaabta si aad u hubiso; kuwa la xaqiijiyay waxaa la furi karaa in xalka ah tallaabo kasta.

Calculate the following derivatives:
1. \(\frac{d}{dx}\text{ln}(5{x}^{3}-2)\)
2. \(\frac{d}{dx}{(\text{ln}(3x))}^{2}\)
Jawaabta muuji
We need to apply the chain rule in both cases.
1. \(\frac{d}{dx}\text{ln}(5{x}^{3}-2)=\frac{15{x}^{2}}{5{x}^{3}-2}\)
2. \(\frac{d}{dx}{(\text{ln}(3x))}^{2}=\frac{2(\text{ln}(3x))\cdot 3}{3x}=\frac{2(\text{ln}(3x))}{x}\)
Calculate the following derivatives:
1. \(\frac{d}{dx}\text{ln}(2{x}^{2}+x)\)
2. \(\frac{d}{dx}{(\text{ln}({x}^{3}))}^{2}\)
Jawaabta muuji
1. \(\frac{d}{dx}\text{ln}(2{x}^{2}+x)=\frac{4x+1}{2{x}^{2}+x}\)
2. \(\frac{d}{dx}{(\text{ln}({x}^{3}))}^{2}=\frac{6\ \text{ln}({x}^{3})}{x}\)
Calculate the integral \(\int \frac{x}{{x}^{2}+4}dx.\)

Jawaabta muuji

Using \(u\)-substitution, let \(u={x}^{2}+4.\) Then \(du=2x\ dx\) and we have
\[\int \frac{x}{{x}^{2}+4}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\text{ln}\ |u|+C=\frac{1}{2}\text{ln}\ |{x}^{2}+4|+C=\frac{1}{2}\text{ln}({x}^{2}+4)+C.\]
Calculate the integral \(\int \frac{{x}^{2}}{{x}^{3}+6}dx.\)

Jawaabta muuji

\(\int \frac{{x}^{2}}{{x}^{3}+6}dx=\frac{1}{3}\text{ln}\ |{x}^{3}+6|+C\)
Use properties of logarithms to simplify the following expression into a single logarithm:
\[\text{ln}\ 9-2\ \text{ln}\ 3+\text{ln}(\frac{1}{3}).\]

Jawaabta muuji

We have
\[\text{ln}\ 9-2\ \text{ln}\ 3+\text{ln}(\frac{1}{3})=\text{ln}({3}^{2})-2\ \text{ln}\ 3+\text{ln}({3}^{-1})=2\ \text{ln}\ 3-2\ \text{ln}\ 3-\text{ln}\ 3=\text{-}\text{ln}\ 3.\]
Use properties of logarithms to simplify the following expression into a single logarithm:
\[\text{ln}\ 8-\text{ln}\ 2-\text{ln}(\frac{1}{4}).\]

Jawaabta muuji

\(4\ \text{ln}\ 2\)
Evaluate the following derivatives:
1. \(\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}\)
2. \(\frac{d}{dx}{e}^{3{x}^{2}}\)
Jawaabta muuji
We apply the chain rule as necessary.
1. \(\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}=\frac{d}{dt}{e}^{3t+{t}^{2}}={e}^{3t+{t}^{2}}(3+2t)\)
2. \(\frac{d}{dx}{e}^{3{x}^{2}}={e}^{3{x}^{2}}6x\)
Evaluate the following derivatives:
1. \(\frac{d}{dx}(\frac{{e}^{{x}^{2}}}{{e}^{5x}})\)
2. \(\frac{d}{dt}{({e}^{2t})}^{3}\)
Jawaabta muuji
1. \(\frac{d}{dx}(\frac{{e}^{{x}^{2}}}{{e}^{5x}})={e}^{{x}^{2}-5x}(2x-5)\)
2. \(\frac{d}{dt}{({e}^{2t})}^{3}=6{e}^{6t}\)
Evaluate the following integral: \(\int 2x{e}^{\text{-}{x}^{2}}dx.\)

Jawaabta muuji

Using \(u\)-substitution, let \(u=\text{-}{x}^{2}.\) Then \(du=-2x\ dx,\) and we have
\[\int 2x{e}^{\text{-}{x}^{2}}dx=\text{-}\int {e}^{u}du=\text{-}{e}^{u}+C=\text{-}{e}^{\text{-}{x}^{2}}+C.\]
Evaluate the following integral: \(\int \frac{4}{{e}^{3x}}dx.\)

Jawaabta muuji

\(\int \frac{4}{{e}^{3x}}dx=-\frac{4}{3}{e}^{-3x}+C\)
Evaluate the following derivatives:
1. \(\frac{d}{dt}({4}^{t}\cdot {2}^{{t}^{2}})\)
2. \(\frac{d}{dx}{\text{log}}_{8}(7{x}^{2}+4)\)
Jawaabta muuji
We need to apply the chain rule as necessary.
1. \(\frac{d}{dt}({4}^{t}\cdot {2}^{{t}^{2}})=\frac{d}{dt}({2}^{2t}\cdot {2}^{{t}^{2}})=\frac{d}{dt}({2}^{2t+{t}^{2}})={2}^{2t+{t}^{2}}\text{ln}(2)(2+2t)\)
2. \(\frac{d}{dx}{\text{log}}_{8}(7{x}^{2}+4)=\frac{1}{(7{x}^{2}+4)(\text{ln}\ 8)}(14x)\)
Evaluate the following derivatives:
1. \(\frac{d}{dt}\ {4}^{{t}^{4}}\)
2. \(\frac{d}{dx}{\text{log}}_{3}(\sqrt{{x}^{2}+1})\)
Jawaabta muuji
1. \(\frac{d}{dt}{4}^{{t}^{4}}={4}^{{t}^{4}}(\text{ln}\ 4)(4{t}^{3})\)
2. \(\frac{d}{dx}{\text{log}}_{3}(\sqrt{{x}^{2}+1})=\frac{x}{(\text{ln}\ 3)({x}^{2}+1)}\)
Evaluate the following integral: \(\int \frac{3}{{2}^{3x}}dx.\)

Jawaabta muuji

Use \(u\text{-substitution}\) and let \(u=-3x.\) Then \(du=-3dx\) and we have
\[\int \frac{3}{{2}^{3x}}dx=\int 3\cdot {2}^{-3x}dx=\text{-}\int {2}^{u}du=-\frac{1}{\text{ln}\ 2}{2}^{u}+C=-\frac{1}{\text{ln}\ 2}{2}^{-3x}+C.\]
Evaluate the following integral: \(\int {x}^{2}{2}^{{x}^{3}}dx.\)

Jawaabta muuji

\(\int {x}^{2}{2}^{{x}^{3}}dx=\frac{1}{3\ \text{ln}\ 2}{2}^{{x}^{3}}+C\)
\(y=\text{ln}(2x)\)

Jawaabta muuji

\(\frac{1}{x}\)
\(y=\text{ln}(2x+1)\)
\(y=\frac{1}{\text{ln}\ x}\)

Jawaabta muuji

\(-\frac{1}{x{(\text{ln}\ x)}^{2}}\)
\(\int \frac{dt}{3t}\)
\(\int \frac{dx}{1+x}\)

Jawaabta muuji

\(\text{ln}(x+1)+C\)
[T] \(y=\frac{\text{ln}(x)}{x}\)
[T] \(y=x\ \text{ln}(x)\)

Jawaabta muuji

\(\text{ln}(x)+1\)
[T] \(y={\text{log}}_{10}x\)
[T] \(y=\text{ln}(\text{sin}\ x)\)

Jawaabta muuji

\(\text{cot}(x)\)
[T] \(y=\text{ln}(\text{ln}\ x)\)
[T] \(y=7\ \text{ln}(4x)\)

Jawaabta muuji

\(\frac{7}{x}\)
[T] \(y=\text{ln}({(4x)}^{7})\)
[T] \(y=\text{ln}(\text{tan}\ x)\)

Jawaabta muuji

\(\text{csc}(x)\text{sec}\ x\)
[T] \(y=\text{ln}(\text{tan}(3x))\)
[T] \(y=\text{ln}({\text{cos}}^{2}x)\)

Jawaabta muuji

\(-2\ \text{tan}\ x\)
\({\int }_{0}^{1}\frac{dx}{3+x}\)
\({\int }_{0}^{1}\frac{dt}{3+2t}\)

Jawaabta muuji

\(\frac{1}{2}\text{ln}(\frac{5}{3})\)
\({\int }_{0}^{2}\frac{x\ dx}{{x}^{2}+1}\)
\({\int }_{0}^{2}\frac{{x}^{3}dx}{{x}^{2}+1}\)

Jawaabta muuji

\(2-\frac{1}{2}\text{ln}(5)\)
\({\int }_{2}^{e}\frac{dx}{x\ \text{ln}\ x}\)
\({\int }_{2}^{e}\frac{dx}{{x\ (\text{ln}\ x)}^{2}}\)

Jawaabta muuji

\(\frac{1}{\text{ln}(2)}-1\)
\(\int \frac{\text{cos}\ x\ dx}{\text{sin}\ x}\)
\({\int }_{0}^{\pi \text{/}4}\text{tan}\ x\ dx\)

Jawaabta muuji

\(\frac{1}{2}\text{ln}(2)\)
\(\int \text{cot}(3x)dx\)
\(\int \frac{{(\text{ln}\ x)}^{2}dx}{x}\)

Jawaabta muuji

\(\frac{1}{3}{(\text{ln}\ x)}^{3}\)
\(y=\sqrt{{x}^{2}+1}\)

Ku hay shaqadaada

A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.

Ku soo biir Ku soo gal

Calaamadaha la isticmaalay halkan

Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.

Sida loo: Integrals, Exponential Functions, and Logarithms

Write the definition of the natural logarithm as an integral.
Recognize the derivative of the natural logarithm.
Integrate functions involving the natural logarithmic function.
Define the number
Recognize the derivative and integral of the exponential function.
Prove properties of logarithms and exponential functions using integrals.
Express general logarithmic and exponential functions in terms of natural logarithms and exponentials.
The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally. This section develops the concepts in a mathematically rigorous way.

Su'aalaha dadku waydiiyaan

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.

Qaybaha ee boggaan waxaa la beddelay ka OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo. Sababtoo ah asalka waxaa loo oggolaaday in aan loo isticmaalin ganacsi, fasalkan waa bilaash, waxba ma iibiyo, iyo waxaa lagu wadaagayaa liisan la mid ah.

In ka badan 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