Calculus Examples

$\int_{- \infty}^{\infty} x e^{1 - x^{2}} d x$

Step 1

Split the integral at $0$ and write as a sum of limits.

$lim_{t \to - \infty} \int_{t}^{0} x e^{1 - x^{2}} d x + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 2

Let $u = 1 - x^{2}$ . Then $d u = - 2 x d x$ , so $- \frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = 1 - x^{2}$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $1 - x^{2}$ .

$\frac{d}{d x} [1 - x^{2}]$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $1 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x^{2}]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [- x^{2}]$

Step 2.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- x^{2}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- x^{2}]$ .

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Step 2.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$0 - \frac{d}{d x} [x^{2}]$

Step 2.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$0 - (2 x)$

Step 2.1.3.3

Multiply $2$ by $- 1$ .

$0 - 2 x$

Step 2.1.4

Subtract $2 x$ from $0$ .

$- 2 x$

Step 2.2

Substitute the lower limit in for $x$ in $u = 1 - x^{2}$ .

$u_{lower} = 1 - t^{2}$

Step 2.3

Substitute the upper limit in for $x$ in $u = 1 - x^{2}$ .

$u_{upper} = 1 - 0^{2}$

Step 2.4

Simplify.

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Step 2.4.1

Simplify each term.

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Step 2.4.1.1

Raising $0$ to any positive power yields $0$ .

$u_{upper} = 1 - 0$

Step 2.4.1.2

Multiply $- 1$ by $0$ .

$u_{upper} = 1 + 0$

Step 2.4.2

Add $1$ and $0$ .

$u_{upper} = 1$

Step 2.5

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

$u_{lower} = 1 - t^{2}$

$u_{upper} = 1$

Step 2.6

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

$lim_{t \to - \infty} \int_{1 - t^{2}}^{1} e^{u} \frac{1}{- 2} d u + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 3

Simplify.

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Step 3.1

Move the negative in front of the fraction.

$lim_{t \to - \infty} \int_{1 - t^{2}}^{1} e^{u} (- \frac{1}{2}) d u + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 3.2

Combine $e^{u}$ and $\frac{1}{2}$ .

$lim_{t \to - \infty} \int_{1 - t^{2}}^{1} - \frac{e^{u}}{2} d u + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 4

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

$lim_{t \to - \infty} - \int_{1 - t^{2}}^{1} \frac{e^{u}}{2} d u + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 5

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$lim_{t \to - \infty} - (\frac{1}{2} \int_{1 - t^{2}}^{1} e^{u} d u) + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 6

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

$lim_{t \to - \infty} - \frac{1}{2} e^{u}]_{1 - t^{2}}^{1} + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 7

Combine $e^{u}]_{1 - t^{2}}^{1}$ and $\frac{1}{2}$ .

$lim_{t \to - \infty} - \frac{e^{u}]_{1 - t^{2}}^{1}}{2} + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $e^{u}$ at $1$ and at $1 - t^{2}$ .

$lim_{t \to - \infty} - \frac{(e^{1}) - e^{1 - t^{2}}}{2} + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 8.2

Simplify.

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} \int_{0}^{t} x e^{1 - x^{2}} d x$

Step 9

Let $u = 1 - x^{2}$ . Then $d u = - 2 x d x$ , so $- \frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 9.1

Let $u = 1 - x^{2}$ . Find $\frac{d u}{d x}$ .

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Step 9.1.1

Differentiate $1 - x^{2}$ .

$\frac{d}{d x} [1 - x^{2}]$

Step 9.1.2

Differentiate.

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Step 9.1.2.1

By the Sum Rule, the derivative of $1 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x^{2}]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [- x^{2}]$

Step 9.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- x^{2}]$

Step 9.1.3

Evaluate $\frac{d}{d x} [- x^{2}]$ .

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Step 9.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$0 - \frac{d}{d x} [x^{2}]$

Step 9.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$0 - (2 x)$

Step 9.1.3.3

Multiply $2$ by $- 1$ .

$0 - 2 x$

Step 9.1.4

Subtract $2 x$ from $0$ .

$- 2 x$

Step 9.2

Substitute the lower limit in for $x$ in $u = 1 - x^{2}$ .

$u_{lower} = 1 - 0^{2}$

Step 9.3

Simplify.

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Step 9.3.1

Simplify each term.

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Step 9.3.1.1

Raising $0$ to any positive power yields $0$ .

$u_{lower} = 1 - 0$

Step 9.3.1.2

Multiply $- 1$ by $0$ .

$u_{lower} = 1 + 0$

Step 9.3.2

Add $1$ and $0$ .

$u_{lower} = 1$

Step 9.4

Substitute the upper limit in for $x$ in $u = 1 - x^{2}$ .

$u_{upper} = 1 - t^{2}$

Step 9.5

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

$u_{lower} = 1$

$u_{upper} = 1 - t^{2}$

Step 9.6

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

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} \int_{1}^{1 - t^{2}} e^{u} \frac{1}{- 2} d u$

Step 10

Simplify.

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Step 10.1

Move the negative in front of the fraction.

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} \int_{1}^{1 - t^{2}} e^{u} (- \frac{1}{2}) d u$

Step 10.2

Combine $e^{u}$ and $\frac{1}{2}$ .

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} \int_{1}^{1 - t^{2}} - \frac{e^{u}}{2} d u$

Step 11

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

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} - \int_{1}^{1 - t^{2}} \frac{e^{u}}{2} d u$

Step 12

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} - (\frac{1}{2} \int_{1}^{1 - t^{2}} e^{u} d u)$

Step 13

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

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} - \frac{1}{2} e^{u}]_{1}^{1 - t^{2}}$

Step 14

Combine $e^{u}]_{1}^{1 - t^{2}}$ and $\frac{1}{2}$ .

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} - \frac{e^{u}]_{1}^{1 - t^{2}}}{2}$

Step 15

Substitute and simplify.

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Step 15.1

Evaluate $e^{u}$ at $1 - t^{2}$ and at $1$ .

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} - \frac{(e^{1 - t^{2}}) - e^{1}}{2}$

Step 15.2

Simplify.

$lim_{t \to - \infty} - \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} - \frac{e^{1 - t^{2}} - e}{2}$

Step 16

Evaluate the limits.

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Step 16.1

Evaluate the limit.

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Step 16.1.1

Move the term $- 1$ outside of the limit because it is constant with respect to $t$ .

$- lim_{t \to - \infty} \frac{e - e^{1 - t^{2}}}{2} + lim_{t \to \infty} - \frac{e^{1 - t^{2}} - e}{2}$

Step 16.1.2

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{2} lim_{t \to - \infty} e - e^{1 - t^{2}} + lim_{t \to \infty} - \frac{e^{1 - t^{2}} - e}{2}$

Step 16.1.3

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

$- \frac{1}{2} (lim_{t \to - \infty} e - lim_{t \to - \infty} e^{1 - t^{2}}) + lim_{t \to \infty} - \frac{e^{1 - t^{2}} - e}{2}$

Step 16.1.4

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

$- \frac{1}{2} (e - lim_{t \to - \infty} e^{1 - t^{2}}) + lim_{t \to \infty} - \frac{e^{1 - t^{2}} - e}{2}$

Step 16.2

Since the exponent $1 - t^{2}$ approaches $- \infty$ , the quantity $e^{1 - t^{2}}$ approaches $0$ .

$- \frac{1}{2} (e - 0) + lim_{t \to \infty} - \frac{e^{1 - t^{2}} - e}{2}$

Step 16.3

Evaluate the limit.

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Step 16.3.1

Move the term $- 1$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{2} (e - 0) - lim_{t \to \infty} \frac{e^{1 - t^{2}} - e}{2}$

Step 16.3.2

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{2} (e - 0) - \frac{1}{2} lim_{t \to \infty} e^{1 - t^{2}} - e$

Step 16.3.3

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

$- \frac{1}{2} (e - 0) - \frac{1}{2} (lim_{t \to \infty} e^{1 - t^{2}} - lim_{t \to \infty} e)$

Step 16.4

Since the exponent $1 - t^{2}$ approaches $- \infty$ , the quantity $e^{1 - t^{2}}$ approaches $0$ .

$- \frac{1}{2} (e - 0) - \frac{1}{2} (0 - lim_{t \to \infty} e)$

Step 16.5

Evaluate the limit.

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Step 16.5.1

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

$- \frac{1}{2} (e - 0) - \frac{1}{2} (0 - e)$

Step 16.5.2

Simplify the answer.

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Step 16.5.2.1

Simplify each term.

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Step 16.5.2.1.1

Subtract $0$ from $e$ .

$- \frac{1}{2} e - \frac{1}{2} (0 - e)$

Step 16.5.2.1.2

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

$- \frac{e}{2} - \frac{1}{2} (0 - e)$

Step 16.5.2.1.3

Subtract $e$ from $0$ .

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

Step 16.5.2.1.4

Multiply $- \frac{1}{2} (- e)$ .

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Step 16.5.2.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 16.5.2.1.4.2

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

$- \frac{e}{2} + \frac{1}{2} e$

Step 16.5.2.1.4.3

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

$- \frac{e}{2} + \frac{e}{2}$

Step 16.5.2.2

Combine the numerators over the common denominator.

$\frac{- e + e}{2}$

Step 16.5.2.3

Add $- e$ and $e$ .

$\frac{0}{2}$

Step 16.5.2.4

Divide $0$ by $2$ .

$0$