Calculus Examples

$\int_{2}^{3} 0.1 e^{- 0.1 a} + \frac{2}{a} d a$

Step 1

Split the single integral into multiple integrals.

$\int_{2}^{3} 0.1 e^{- 0.1 a} d a + \int_{2}^{3} \frac{2}{a} d a$

Step 2

Since $0.1$ is constant with respect to $a$ , move $0.1$ out of the integral.

$0.1 \int_{2}^{3} e^{- 0.1 a} d a + \int_{2}^{3} \frac{2}{a} d a$

Step 3

Let $u = - 0.1 a$ . Then $d u = - 0.1 d a$ , so $\frac{1}{- 0.1} d u = d a$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.1

Let $u = - 0.1 a$ . Find $\frac{d u}{d a}$ .

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Step 3.1.1

Differentiate $- 0.1 a$ .

$\frac{d}{d a} [- 0.1 a]$

Step 3.1.2

Since $- 0.1$ is constant with respect to $a$ , the derivative of $- 0.1 a$ with respect to $a$ is $- 0.1 \frac{d}{d a} [a]$ .

$- 0.1 \frac{d}{d a} [a]$

Step 3.1.3

Differentiate using the Power Rule which states that $\frac{d}{d a} [a^{n}]$ is $n a^{n - 1}$ where $n = 1$ .

$- 0.1 \cdot 1$

Step 3.1.4

Multiply $- 0.1$ by $1$ .

$- 0.1$

Step 3.2

Substitute the lower limit in for $a$ in $u = - 0.1 a$ .

$u_{lower} = - 0.1 \cdot 2$

Step 3.3

Multiply $- 0.1$ by $2$ .

$u_{lower} = - 0.2$

Step 3.4

Substitute the upper limit in for $a$ in $u = - 0.1 a$ .

$u_{upper} = - 0.1 \cdot 3$

Step 3.5

Multiply $- 0.1$ by $3$ .

$u_{upper} = - 0.3$

Step 3.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = - 0.2$

$u_{upper} = - 0.3$

Step 3.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$0.1 \int_{- 0.2}^{- 0.3} e^{u} \frac{1}{- 0.1} d u + \int_{2}^{3} \frac{2}{a} d a$

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

$0.1 \int_{- 0.2}^{- 0.3} e^{u} (- \frac{1}{0.1}) d u + \int_{2}^{3} \frac{2}{a} d a$

Step 4.2

Combine $e^{u}$ and $\frac{1}{0.1}$ .

$0.1 \int_{- 0.2}^{- 0.3} - \frac{e^{u}}{0.1} d u + \int_{2}^{3} \frac{2}{a} d a$

Step 5

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$0.1 (- \int_{- 0.2}^{- 0.3} \frac{e^{u}}{0.1} d u) + \int_{2}^{3} \frac{2}{a} d a$

Step 6

Multiply $- 1$ by $0.1$ .

$- 0.1 \int_{- 0.2}^{- 0.3} \frac{e^{u}}{0.1} d u + \int_{2}^{3} \frac{2}{a} d a$

Step 7

Since $\frac{1}{0.1}$ is constant with respect to $u$ , move $\frac{1}{0.1}$ out of the integral.

$- 0.1 (\frac{1}{0.1} \int_{- 0.2}^{- 0.3} e^{u} d u) + \int_{2}^{3} \frac{2}{a} d a$

Step 8

Simplify.

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Step 8.1

Combine $\frac{1}{0.1}$ and $- 0.1$ .

$\frac{- 0.1}{0.1} \int_{- 0.2}^{- 0.3} e^{u} d u + \int_{2}^{3} \frac{2}{a} d a$

Step 8.2

Cancel the common factor of $- 0.1$ and $0.1$ .

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Step 8.2.1

Rewrite $- 0.1$ as $- 1 (0.1)$ .

$\frac{- 1 (0.1)}{0.1} \int_{- 0.2}^{- 0.3} e^{u} d u + \int_{2}^{3} \frac{2}{a} d a$

Step 8.2.2

Cancel the common factor.

$\frac{- 1 \cdot 0.1}{0.1} \int_{- 0.2}^{- 0.3} e^{u} d u + \int_{2}^{3} \frac{2}{a} d a$

Step 8.2.3

Divide $- 1$ by $1$ .

$- \int_{- 0.2}^{- 0.3} e^{u} d u + \int_{2}^{3} \frac{2}{a} d a$

Step 9

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$- (e^{u}]_{- 0.2}^{- 0.3}) + \int_{2}^{3} \frac{2}{a} d a$

Step 10

Since $2$ is constant with respect to $a$ , move $2$ out of the integral.

$- (e^{u}]_{- 0.2}^{- 0.3}) + 2 \int_{2}^{3} \frac{1}{a} d a$

Step 11

The integral of $\frac{1}{a}$ with respect to $a$ is $\ln (| a |)$ .

$- (e^{u}]_{- 0.2}^{- 0.3}) + 2 (\ln (| a |)]_{2}^{3})$

Step 12

Substitute and simplify.

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Step 12.1

Evaluate $e^{u}$ at $- 0.3$ and at $- 0.2$ .

$- ((e^{- 0.3}) - e^{- 0.2}) + 2 (\ln (| a |)]_{2}^{3})$

Step 12.2

Evaluate $\ln (| a |)$ at $3$ and at $2$ .

$- ((e^{- 0.3}) - e^{- 0.2}) + 2 ((\ln (| 3 |)) - \ln (| 2 |))$

Step 12.3

Remove parentheses.

$- (e^{- 0.3} - e^{- 0.2}) + 2 (\ln (| 3 |) - \ln (| 2 |))$

Step 13

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$- (e^{- 0.3} - e^{- 0.2}) + 2 \ln (\frac{| 3 |}{| 2 |})$

Step 14

Simplify.

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Step 14.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- (e^{- 0.3} - \frac{1}{e^{0.2}}) + 2 \ln (\frac{| 3 |}{| 2 |})$

Step 14.2

Apply the distributive property.

$- e^{- 0.3} - - \frac{1}{e^{0.2}} + 2 \ln (\frac{| 3 |}{| 2 |})$

Step 14.3

Multiply $- - \frac{1}{e^{0.2}}$ .

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Step 14.3.1

Multiply $- 1$ by $- 1$ .

$- e^{- 0.3} + 1 \frac{1}{e^{0.2}} + 2 \ln (\frac{| 3 |}{| 2 |})$

Step 14.3.2

Multiply $\frac{1}{e^{0.2}}$ by $1$ .

$- e^{- 0.3} + \frac{1}{e^{0.2}} + 2 \ln (\frac{| 3 |}{| 2 |})$

Step 14.4

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$- e^{- 0.3} + \frac{1}{e^{0.2}} + 2 \ln (\frac{3}{| 2 |})$

Step 14.5

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$- e^{- 0.3} + \frac{1}{e^{0.2}} + 2 \ln (\frac{3}{2})$

Step 14.6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{e^{0.3}} + \frac{1}{e^{0.2}} + 2 \ln (\frac{3}{2})$

Step 15

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{e^{0.3}} + \frac{1}{e^{0.2}} + 2 \ln (\frac{3}{2})$

Decimal Form:

$0.88884274 \dots$

Step 16