Calculus Examples

$\int_{0}^{\infty} x e^{- x} d x$

Step 1

Anything raised to $0$ is $1$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} e^{- t} - 1 \cdot 1$

Step 2

Multiply $- 1$ by $1$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} e^{- t} - 1$

Step 3

Evaluate the limit.

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Step 3.1

Rewrite.

$\int_{0}^{\infty} x e^{- x} d x$

Step 3.2

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

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

Step 3.3

Write the integral as a limit as $t$ approaches $\infty$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} \int_{0}^{t} - e^{- x} d x$

Step 3.4

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

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} - \int_{0}^{t} e^{- x} d x$

Step 3.5

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

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Step 3.5.1

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

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Step 3.5.1.1

Differentiate $- x$ .

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

Step 3.5.1.2

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 3.5.1.3

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 3.5.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 3.5.2

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

$u_{lower} = - 0$

Step 3.5.3

Multiply $- 1$ by $0$ .

$u_{lower} = 0$

Step 3.5.4

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

$u_{upper} = - t$

Step 3.5.5

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

$u_{lower} = 0$

$u_{upper} = - t$

Step 3.5.6

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

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} - \int_{0}^{- t} - e^{u} d u$

Step 3.6

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

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} - - \int_{0}^{- t} e^{u} d u$

Step 3.7

Simplify.

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Step 3.7.1

Multiply $- 1$ by $- 1$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} 1 \int_{0}^{- t} e^{u} d u$

Step 3.7.2

Multiply $\int_{0}^{- t} e^{u} d u$ by $1$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} \int_{0}^{- t} e^{u} d u$

Step 3.8

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

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} e^{u}]_{0}^{- t}$

Step 3.9

Evaluate the limit.

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Step 3.9.1

Simplify the expression.

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Step 3.9.1.1

Evaluate $e^{u}$ at $- t$ and at $0$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} (e^{- t}) - e^{0}$

Step 3.9.1.2

Anything raised to $0$ is $1$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} e^{- t} - 1 \cdot 1$

Step 3.9.1.3

Multiply $- 1$ by $1$ .

$x (- e^{- x})]_{0}^{\infty} - lim_{t \to \infty} e^{- t} - 1$

Step 3.9.2

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $\infty$ .

$x \cdot - 1 e^{- x}]_{0}^{\infty} - (lim_{t \to \infty} e^{- t} - lim_{t \to \infty} 1)$

Step 3.10

Since the exponent $- t$ approaches $- \infty$ , the quantity $e^{- t}$ approaches $0$ .

$x \cdot - 1 e^{- x}]_{0}^{\infty} - (0 - lim_{t \to \infty} 1)$

Step 3.11

Evaluate the limit.

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Step 3.11.1

Evaluate the limit of $1$ which is constant as $t$ approaches $\infty$ .

$x \cdot - 1 e^{- x}]_{0}^{\infty} - (0 - 1 \cdot 1)$

Step 3.11.2

Simplify the answer.

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Step 3.11.2.1

Subtract $1 \cdot 1$ from $0$ .

$x \cdot - 1 e^{- x}]_{0}^{\infty} - (- 1 \cdot 1)$

Step 3.11.2.2

Simplify each term.

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Step 3.11.2.2.1

Move $- 1$ to the left of $x$ .

$- 1 \cdot x e^{- x}]_{0}^{\infty} - (- 1 \cdot 1)$

Step 3.11.2.2.2

Rewrite $- 1 x$ as $- x$ .

$- x e^{- x}]_{0}^{\infty} - (- 1 \cdot 1)$

Step 3.11.2.2.3

Multiply $- (- 1 \cdot 1)$ .

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Step 3.11.2.2.3.1

Multiply $- 1$ by $1$ .

$- x e^{- x}]_{0}^{\infty} - - 1$

Step 3.11.2.2.3.2

Multiply $- 1$ by $- 1$ .

$- x e^{- x}]_{0}^{\infty} + 1$