Definite Integrals of Logarithmic Functions

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Question 1 of 6

Find the definite integral

\int_{0}^{2} \frac{x}{x^{2} + 1}

$\int_{0}^{2} \frac{x}{x^2+1}$

1. $2 ln 5$ $2 ln 5$
2. $\frac{1}{5}$ $1/5$
3. $- \frac{1}{2} ln 5$ $-1/2 ln 5$
4. $\frac{1}{2} ln 5$ $1/2 ln 5$

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Integral of a Basic Fraction

\int \frac{1}{x} d x = {log}_{e} x + c

$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$

General Form Integral of a Fraction

\int \frac{f^{'} (x)}{f (x)} d x = {log}_{e} [f (x)] + c

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\frac{x}{\color{#CC0000}{x^2+1}}$	$=$ $=$	$\frac{x}{\color{#CC0000}{2x}}$	Differentiate $x^{2} + 1$ $x^2+1$

	$=$ $=$	$\frac{1}{2}$ $1/2$	$\frac{x}{x} = 1$ $x/x=1$

Use

\frac{1}{2}

$1/2$ as a constant to balance the integral.

$f (x)$ $f(x)$	$=$ $=$	$x^{2} + 1$ $x^2+1$
$f' (x)$ $f'(x)$	$=$ $=$	$2 x$ $2x$

	$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$

$=$ $=$	$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{2x}}{\color{#D800AD}{x^2+1}}dx$	Substitute known values

$=$ $=$	${[\frac{1}{2} ln (x^{2} + 1)]}_{0}^{2}$ $[1/2 ln (x^2+1)]_0^2$

Finally, get the difference of the upper and lower limits substituted to the integral as

x

$x$ .

	${[\frac{1}{2} ln (x^{2} + 1)]}_{0}^{2}$ $[1/2 ln (x^2+1)]_0^2$

$=$	$1/2[ln ((2)^2+1)-ln((0)^2+1)]$	Substitute the limits

$=$	$1/2[ln 5-ln1]$	Evaluate

$=$	$1/2 ln 5$

$1/2 ln 5$

Question 2 of 6

Find the definite integral

$\int_{1}^{7} \frac{x^2}{x^3+2}$

Round your answer to two decimal places

(1.58)

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Integral of a Basic Fraction

$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$

General Form Integral of a Fraction

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\frac{x^2}{\color{#CC0000}{x^3+2}}$	$=$	$\frac{x^2}{\color{#CC0000}{3x^2}}$	Differentiate $x^3+2$

	$=$	$1/3$	$x^2/x^2=1$

Use

$1/3$ as a constant to balance the integral.

$f(x)$	$=$	$x^3+2$
$f'(x)$	$=$	$3x^2$

	$\frac{1}{3}\int_{1}^{7} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$

$=$	$\frac{1}{3}\int_{1}^{7} \frac{\color{#9a00c7}{3x^2}}{\color{#D800AD}{x^3+2}}dx$	Substitute known values

$=$	$[1/3 ln (x^3+2)]_1^7$

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$[1/3 ln (x^3+2)]_1^7$

$=$	$1/3[ln ((7)^3+2)-ln((1)^3+2)]$	Substitute the limits

$=$	$1/3[ln 347-ln 3]$	Evaluate

$=$	$1/3[4.7507]$	Compute using calculator

$=$	$1.58$	Rounded to two decimal places

$1.58$

Question 3 of 6

Find the definite integral

$\int_{1}^{3} \frac{1}{2x-1}$

1. $1/2 ln 5$
2. $2 ln 5$
3. $1/5$
4. $-1/2 ln 5$

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Integral of a Basic Fraction

$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$

General Form Integral of a Fraction

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\frac{1}{\color{#CC0000}{2x-1}}$

$=$

$\frac{1}{\color{#CC0000}{2}}$

Differentiate

$2x-1$

Use

$1/2$ as a constant to balance the integral.

$f(x)$	$=$	$2x-1$
$f'(x)$	$=$	$2$

	$\frac{1}{2}\int_{1}^{3} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$

$=$	$\frac{1}{2}\int_{1}^{3} \frac{\color{#9a00c7}{2}}{\color{#D800AD}{2x-1}}dx$	Substitute known values

$=$	$[1/2 ln (2x-1)]_1^3$

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$[1/2 ln (2x-1)]_1^3$

$=$	$1/2[ln (2(3)-1)-ln(2(1)-1)]$	Substitute the limits

$=$	$1/2[ln 5-0]$	Evaluate

$=$	$1/2 ln 5$

$1/2 ln 5$

Question 4 of 6

Find the area under the curve

$y=4/x$

Round your answer to three decimal places

(4.394) $\text(units)^2$

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Integral of a Basic Fraction

$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$

General Form Integral of a Fraction

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\frac{4}{\color{#CC0000}{x}}$	$=$	$\frac{4}{\color{#CC0000}{4}}$	Differentiate $x^3+2$

	$=$	$4$	$4/1=4$

We can see the limits indicated in the graph

$\text(Upper Limit)$	$=$	$3$
$\text(Lower Limit)$	$=$	$1$

Use

$4$ as a constant to balance the integral.

$f(x)$	$=$	$x$
$f'(x)$	$=$	$1$

	$4\int_{1}^{3} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$

$=$	$4\int_{1}^{7} \frac{\color{#9a00c7}{1}}{\color{#D800AD}{x}}dx$	Substitute known values

$=$	$4[ln x]_1^3$

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$4[ln x]_1^3$
$=$	$4[ln (3)-ln (1)]$	Substitute the limits
$=$	$4[ln 3-ln 1]$	Evaluate
$=$	$4 ln 3$
$=$	$4.394$	Compute using calculator

$4.394 \text(units)^2$

Question 5 of 6

Find the area under the curve

$y=x/(x^2+1)$ going to the x-axis

Round your answer to two decimal places

(0.80, 0.8) $\text(units)^2$

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Integral of a Basic Fraction

$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$

General Form Integral of a Fraction

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\frac{x}{\color{#CC0000}{x^2+1}}$	$=$	$\frac{x}{\color{#CC0000}{2x}}$	Differentiate $x^3+2$

	$=$	$1/2$	$x/x=1$

We can see the limits indicated in the graph

$\text(Upper Limit)$	$=$	$2$
$\text(Lower Limit)$	$=$	$0$

Use

$1/2$ as a constant to balance the integral.

$f(x)$	$=$	$x^2+1$
$f'(x)$	$=$	$2x$

	$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$

$=$	$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{2x}}{\color{#D800AD}{x^2+1}}dx$	Substitute known values

$=$	$1/2[ln (x^2+1)]_0^2$

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$1/2[ln (x^2+1)]_0^2$

$=$	$1/2[ln ((2)^2+1)-ln ((0)^2+1)]$	Substitute the limits

$=$	$1/2[ln 5-ln 1]$	Evaluate

$=$	$1/2 ln 5$

$=$	$0.80$	Compute using calculator

$0.80 \text(units)^2$

Question 6 of 6

6. Question
Find the area under the curve $y=lnx$ going to the x-axis
- 1. $3ln 3-2 \text(units)^2$
- 2. $2 \text(units)^2$
- 3. $3ln 3+2 \text(units)^2$
- 4. $3ln 3 \text(units)^2$
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Integral of a Basic Fraction

$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$

General Form Integral of a Fraction

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

Since we cannot integrate $ln x$ , we can find the value of $x$ and integrate it.

$y$ $=$ $ln x$

$e^y$ $=$ $e^(ln x)$ Insert $e$ as base to both sides

$e^y$ $=$ $x$ $e^(ln x)=x$

$x$ $=$ $e^y$

Therefore, the formula that we can use to find the area is $int e^y dy$

We also need to find the limits for $y$

$\text(Lower Limit)$ $=$ $0$

We can see on the graph that the upper limit on the x-axis is $3$

Substitute this value to the equation to find the upper limit on the y-axis

$y$ $=$ $ln x$

$y$ $=$ $ln 3$ $x=3$

$\text(Upper Limit)$ $=$ $ln 3$

Next, integrate and get the difference of the upper and lower limits substituted to the
integral as $x$ .

$int_0^(ln3) e^y dy$

$=$ $[e^y]_0^(ln3)$ Integrate

$=$ $e^((ln3))-e^((0))$ Substitute the limits

$=$ $3-1$ Anything raised to $0$ is $1$

$=$ $2$

The area of the curve going to the y-axis is $2 \text(units)^2$

To find the area of the curve going to the x-axis, find the area of the whole rectangle and subtract the area of the curve going to the y-axis.

$\text(Area)$ $=$ $\text(length)times\text(width)$

$=$ $ln 3times3$

$=$ $3ln 3-$ $2$

$3ln 3-2 \text(units)^2$

Definite Integrals of Logarithmic Functions

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Quiz summary

Information

1. Question

Hint

Great Work!

Integral of a Basic Fraction

General Form Integral of a Fraction

2. Question

Hint

Correct!

Integral of a Basic Fraction

General Form Integral of a Fraction

3. Question

Hint

Keep Going!

Integral of a Basic Fraction

General Form Integral of a Fraction

4. Question

Hint

Fantastic!

Integral of a Basic Fraction

General Form Integral of a Fraction

5. Question

Hint

Excellent!

Integral of a Basic Fraction

General Form Integral of a Fraction

6. Question

Hint

Nice Job!

Integral of a Basic Fraction

General Form Integral of a Fraction

Quizzes

$y$	$=$	$ln x$
$e^y$	$=$	$e^(ln x)$	Insert $e$ as base to both sides
$e^y$	$=$	$x$	$e^(ln x)=x$
$x$	$=$	$e^y$

	$int_0^(ln3) e^y dy$
$=$	$[e^y]_0^(ln3)$	Integrate
$=$	$e^((ln3))-e^((0))$	Substitute the limits
$=$	$3-1$	Anything raised to $0$ is $1$
$=$	$2$

$\text(Area)$	$=$	$\text(length)times\text(width)$
	$=$	$ln 3times3$
	$=$	$3ln 3-$ $2$