Calculus Examples

$\int_{0}^{\ln (2)} x e^{- x} d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = e^{- x}$ .

$x (- e^{- x})]_{0}^{\ln (2)} - \int_{0}^{\ln (2)} - e^{- x} d x$

Step 2

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

$x (- e^{- x})]_{0}^{\ln (2)} - - \int_{0}^{\ln (2)} e^{- x} d x$

Step 3

Simplify.

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Step 3.1

Multiply $- 1$ by $- 1$ .

$x (- e^{- x})]_{0}^{\ln (2)} + 1 \int_{0}^{\ln (2)} e^{- x} d x$

Step 3.2

Multiply $\int_{0}^{\ln (2)} e^{- x} d x$ by $1$ .

$x (- e^{- x})]_{0}^{\ln (2)} + \int_{0}^{\ln (2)} e^{- x} d x$

Step 4

Let $u = - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

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

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Step 4.1.1

Differentiate $- x$ .

$\frac{d}{d x} [- x]$

Step 4.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$- \frac{d}{d x} [x]$

Step 4.1.3

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

$- 1 \cdot 1$

Step 4.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 4.2

Substitute the lower limit in for $x$ in $u = - x$ .

$u_{lower} = - 0$

Step 4.3

Multiply $- 1$ by $0$ .

$u_{lower} = 0$

Step 4.4

Substitute the upper limit in for $x$ in $u = - x$ .

$u_{upper} = - \ln (2)$

Step 4.5

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

$u_{lower} = 0$

$u_{upper} = - \ln (2)$

Step 4.6

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

$x (- e^{- x})]_{0}^{\ln (2)} + \int_{0}^{- \ln (2)} - e^{u} d u$

Step 5

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

$x (- e^{- x})]_{0}^{\ln (2)} - \int_{0}^{- \ln (2)} e^{u} d u$

Step 6

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

$x (- e^{- x})]_{0}^{\ln (2)} - (e^{u}]_{0}^{- \ln (2)})$

Step 7

Substitute and simplify.

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Step 7.1

Evaluate $x (- e^{- x})$ at $\ln (2)$ and at $0$ .

$(\ln (2) (- e^{- \ln (2)})) + 0 (- e^{- 0}) - (e^{u}]_{0}^{- \ln (2)})$

Step 7.2

Evaluate $e^{u}$ at $- \ln (2)$ and at $0$ .

$(\ln (2) (- e^{- \ln (2)})) + 0 (- e^{- 0}) - ((e^{- \ln (2)}) - e^{0})$

Step 7.3

Simplify.

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Step 7.3.1

Multiply $- 1$ by $0$ .

$\ln (2) (- e^{- \ln (2)}) + 0 (- e^{0}) - ((e^{- \ln (2)}) - e^{0})$

Step 7.3.2

Anything raised to $0$ is $1$ .

$\ln (2) (- e^{- \ln (2)}) + 0 (- 1 \cdot 1) - ((e^{- \ln (2)}) - e^{0})$

Step 7.3.3

Multiply $- 1$ by $1$ .

$\ln (2) (- e^{- \ln (2)}) + 0 \cdot - 1 - ((e^{- \ln (2)}) - e^{0})$

Step 7.3.4

Multiply $0$ by $- 1$ .

$\ln (2) (- e^{- \ln (2)}) + 0 - ((e^{- \ln (2)}) - e^{0})$

Step 7.3.5

Add $\ln (2) (- e^{- \ln (2)})$ and $0$ .

$\ln (2) (- e^{- \ln (2)}) - ((e^{- \ln (2)}) - e^{0})$

Step 7.3.6

Anything raised to $0$ is $1$ .

$\ln (2) (- e^{- \ln (2)}) - (e^{- \ln (2)} - 1 \cdot 1)$

Step 7.3.7

Multiply $- 1$ by $1$ .

$\ln (2) (- e^{- \ln (2)}) - (e^{- \ln (2)} - 1)$

Step 8

Simplify each term.

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Step 8.1

Simplify $- \ln (2)$ by moving $- 1$ inside the logarithm.

$\ln (2) (- e^{\ln (2^{- 1})}) - (e^{- \ln (2)} - 1)$

Step 8.2

Exponentiation and log are inverse functions.

$\ln (2) (- 2^{- 1}) - (e^{- \ln (2)} - 1)$

Step 8.3

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

$\ln (2) (- \frac{1}{2}) - (e^{- \ln (2)} - 1)$

Step 8.4

Combine $\ln (2)$ and $\frac{1}{2}$ .

$- \frac{\ln (2)}{2} - (e^{- \ln (2)} - 1)$

Step 8.5

Simplify each term.

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Step 8.5.1

Simplify $- \ln (2)$ by moving $- 1$ inside the logarithm.

$- \frac{\ln (2)}{2} - (e^{\ln (2^{- 1})} - 1)$

Step 8.5.2

Exponentiation and log are inverse functions.

$- \frac{\ln (2)}{2} - (2^{- 1} - 1)$

Step 8.5.3

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

$- \frac{\ln (2)}{2} - (\frac{1}{2} - 1)$

Step 8.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \frac{\ln (2)}{2} - (\frac{1}{2} - 1 \cdot \frac{2}{2})$

Step 8.7

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

$- \frac{\ln (2)}{2} - (\frac{1}{2} + \frac{- 1 \cdot 2}{2})$

Step 8.8

Combine the numerators over the common denominator.

$- \frac{\ln (2)}{2} - \frac{1 - 1 \cdot 2}{2}$

Step 8.9

Simplify the numerator.

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Step 8.9.1

Multiply $- 1$ by $2$ .

$- \frac{\ln (2)}{2} - \frac{1 - 2}{2}$

Step 8.9.2

Subtract $2$ from $1$ .

$- \frac{\ln (2)}{2} - \frac{- 1}{2}$

Step 8.10

Move the negative in front of the fraction.

$- \frac{\ln (2)}{2} - - \frac{1}{2}$

Step 8.11

Multiply $- - \frac{1}{2}$ .

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Step 8.11.1

Multiply $- 1$ by $- 1$ .

$- \frac{\ln (2)}{2} + 1 (\frac{1}{2})$

Step 8.11.2

Multiply $\frac{1}{2}$ by $1$ .

$- \frac{\ln (2)}{2} + \frac{1}{2}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$- \frac{\ln (2)}{2} + \frac{1}{2}$

Decimal Form:

$0.15342640 \dots$

Step 10