Calculus Examples

$\int_{0}^{\infty} 2 x e^{x^{2}} d x$

Step 1

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

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

Step 2

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

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

Step 3

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

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Step 3.1

Let $u_{2} = e^{x^{2}}$ . Find $\frac{d u_{2}}{d x}$ .

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Step 3.1.1

Differentiate $e^{x^{2}}$ .

$\frac{d}{d x} [e^{x^{2}}]$

Step 3.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x^{2}$ .

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Step 3.1.2.1

To apply the Chain Rule, set $u_{1}$ as $x^{2}$ .

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

Step 3.1.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 3.1.2.3

Replace all occurrences of $u_{1}$ with $x^{2}$ .

$e^{x^{2}} \frac{d}{d x} [x^{2}]$

Step 3.1.3

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

$e^{x^{2}} (2 x)$

Step 3.1.4

Simplify.

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Step 3.1.4.1

Reorder the factors of $e^{x^{2}} (2 x)$ .

$2 e^{x^{2}} x$

Step 3.1.4.2

Reorder factors in $2 e^{x^{2}} x$ .

$2 x e^{x^{2}}$

Step 3.2

Substitute the lower limit in for $x$ in $u_{2} = e^{x^{2}}$ .

$u_{lower} = e^{0^{2}}$

Step 3.3

Simplify.

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Step 3.3.1

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

$u_{lower} = e^{0}$

Step 3.3.2

Anything raised to $0$ is $1$ .

$u_{lower} = 1$

Step 3.4

Substitute the upper limit in for $x$ in $u_{2} = e^{x^{2}}$ .

$u_{upper} = e^{t^{2}}$

Step 3.5

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

$u_{lower} = 1$

$u_{upper} = e^{t^{2}}$

Step 3.6

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$lim_{t \to \infty} 2 \int_{1}^{e^{t^{2}}} \frac{1}{2} d u_{2}$

Step 4

Apply the constant rule.

$lim_{t \to \infty} 2 (\frac{1}{2} u_{2}]_{1}^{e^{t^{2}}})$

Step 5

Simplify the answer.

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Step 5.1

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

$lim_{t \to \infty} 2 (\frac{u_{2}}{2}]_{1}^{e^{t^{2}}})$

Step 5.2

Evaluate $\frac{u_{2}}{2}$ at $e^{t^{2}}$ and at $1$ .

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

Step 6

Since the function $\frac{e^{t^{2}}}{2} - \frac{1}{2}$ approaches $\infty$ , the positive constant $2$ times the function also approaches $\infty$ .

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Step 6.1

Consider the limit with the constant multiple $2$ removed.

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

Step 6.2

Evaluate the limit.

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Step 6.2.1

Combine the numerators over the common denominator.

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

Step 6.2.2

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

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

Step 6.2.3

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

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

Step 6.3

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

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

Step 6.4

Evaluate the limit.

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Step 6.4.1

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

$\frac{\infty - 1 \cdot 1}{lim_{t \to \infty} 2}$

Step 6.4.2

Evaluate the limit of $2$ which is constant as $t$ approaches $\infty$ .

$\frac{\infty - 1 \cdot 1}{2}$

Step 6.4.3

Simplify the answer.

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Step 6.4.3.1

Simplify the numerator.

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Step 6.4.3.1.1

Multiply $- 1$ by $1$ .

$\frac{\infty - 1}{2}$

Step 6.4.3.1.2

Infinity plus or minus a number is infinity.

$\frac{\infty}{2}$

Step 6.4.3.2

Infinity divided by anything that is finite and non-zero is infinity.

$\infty$