Calculus Examples

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

Step 1

Split the single integral into multiple integrals.

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

Step 2

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

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

Step 3

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

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Step 3.1

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

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Step 3.1.1

Differentiate $- 0.1 y$ .

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

Step 3.1.2

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

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

Step 3.1.3

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ 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 $y$ in $u = - 0.1 y$ .

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

Step 3.3

Multiply $- 0.1$ by $- 3$ .

$u_{lower} = 0.3$

Step 3.4

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

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

Step 3.5

Multiply $- 0.1$ by $- 2$ .

$u_{upper} = 0.2$

Step 3.6

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

$u_{lower} = 0.3$

$u_{upper} = 0.2$

Step 3.7

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

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

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

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

Step 4.2

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

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

Step 5

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

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

Step 6

Multiply $- 1$ by $2$ .

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

Step 7

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

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

Step 8

Simplify.

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Step 8.1

Combine $\frac{1}{0.1}$ and $- 2$ .

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

Step 8.2

Move the negative in front of the fraction.

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

Step 9

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

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

Step 10

Since $3$ is constant with respect to $y$ , move $3$ out of the integral.

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

Step 11

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

$- \frac{2}{0.1} e^{u}]_{0.3}^{0.2} + 3 (\ln (| y |)]_{- 3}^{- 2})$

Step 12

Substitute and simplify.

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Step 12.1

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

$- \frac{2}{0.1} ((e^{0.2}) - e^{0.3}) + 3 (\ln (| y |)]_{- 3}^{- 2})$

Step 12.2

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

$- \frac{2}{0.1} ((e^{0.2}) - e^{0.3}) + 3 ((\ln (| - 2 |)) - \ln (| - 3 |))$

Step 12.3

Remove parentheses.

$- \frac{2}{0.1} (e^{0.2} - e^{0.3}) + 3 (\ln (| - 2 |) - \ln (| - 3 |))$

Step 13

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

$- \frac{2}{0.1} (e^{0.2} - e^{0.3}) + 3 \ln (\frac{| - 2 |}{| - 3 |})$

Step 14

Simplify.

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Step 14.1

Divide $2$ by $0.1$ .

$- 1 \cdot 20 (e^{0.2} - e^{0.3}) + 3 \ln (\frac{| - 2 |}{| - 3 |})$

Step 14.2

Multiply $- 1$ by $20$ .

$- 20 (e^{0.2} - e^{0.3}) + 3 \ln (\frac{| - 2 |}{| - 3 |})$

Step 14.3

Apply the distributive property.

$- 20 e^{0.2} - 20 (- e^{0.3}) + 3 \ln (\frac{| - 2 |}{| - 3 |})$

Step 14.4

Multiply $- 1$ by $- 20$ .

$- 20 e^{0.2} + 20 e^{0.3} + 3 \ln (\frac{| - 2 |}{| - 3 |})$

Step 14.5

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

$- 20 e^{0.2} + 20 e^{0.3} + 3 \ln (\frac{2}{| - 3 |})$

Step 14.6

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

$- 20 e^{0.2} + 20 e^{0.3} + 3 \ln (\frac{2}{3})$

Step 15

The result can be shown in multiple forms.

Exact Form:

$- 20 e^{0.2} + 20 e^{0.3} + 3 \ln (\frac{2}{3})$

Decimal Form:

$1.35272566 \dots$

Step 16