Calculus Examples

$\int_{0}^{1} x e^{- 9 x^{2}} d x$

Step 1

Let $u_{2} = e^{- 9 x^{2}}$ . Then $d u_{2} = - 18 x e^{- 9 x^{2}} d x$ , so $- \frac{1}{18} d u_{2} = x e^{- 9 x^{2}} d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 1.1

Let $u_{2} = e^{- 9 x^{2}}$ . Find $\frac{d u_{2}}{d x}$ .

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Step 1.1.1

Differentiate $e^{- 9 x^{2}}$ .

$\frac{d}{d x} [e^{- 9 x^{2}}]$

Step 1.1.2

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 $g (x) = - 9 x^{2}$ .

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Step 1.1.2.1

To apply the Chain Rule, set $u_{1}$ as $- 9 x^{2}$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.2

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

$e^{u_{1}} \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.3

Replace all occurrences of $u_{1}$ with $- 9 x^{2}$ .

$e^{- 9 x^{2}} \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x^{2}$ with respect to $x$ is $- 9 \frac{d}{d x} [x^{2}]$ .

$e^{- 9 x^{2}} (- 9 \frac{d}{d x} [x^{2}])$

Step 1.1.3.2

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

$e^{- 9 x^{2}} (- 9 (2 x))$

Step 1.1.3.3

Multiply $2$ by $- 9$ .

$e^{- 9 x^{2}} (- 18 x)$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Reorder the factors of $e^{- 9 x^{2}} (- 18 x)$ .

$- 18 e^{- 9 x^{2}} x$

Step 1.1.4.2

Reorder factors in $- 18 e^{- 9 x^{2}} x$ .

$- 18 x e^{- 9 x^{2}}$

Step 1.2

Substitute the lower limit in for $x$ in $u_{2} = e^{- 9 x^{2}}$ .

$u_{lower} = e^{- 9 \cdot 0^{2}}$

Step 1.3

Simplify.

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Step 1.3.1

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

$u_{lower} = e^{- 9 \cdot 0}$

Step 1.3.2

Multiply $- 9$ by $0$ .

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

Step 1.3.3

Anything raised to $0$ is $1$ .

$u_{lower} = 1$

Step 1.4

Substitute the upper limit in for $x$ in $u_{2} = e^{- 9 x^{2}}$ .

$u_{upper} = e^{- 9 \cdot 1^{2}}$

Step 1.5

Simplify.

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Step 1.5.1

One to any power is one.

$u_{upper} = e^{- 9 \cdot 1}$

Step 1.5.2

Multiply $- 9$ by $1$ .

$u_{upper} = e^{- 9}$

Step 1.5.3

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

$u_{upper} = \frac{1}{e^{9}}$

Step 1.6

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

$u_{lower} = 1$

$u_{upper} = \frac{1}{e^{9}}$

Step 1.7

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

$\int_{1}^{\frac{1}{e^{9}}} \frac{1}{- 18} d u_{2}$

Step 2

Move the negative in front of the fraction.

$\int_{1}^{\frac{1}{e^{9}}} - \frac{1}{18} d u_{2}$

Step 3

Apply the constant rule.

$- \frac{1}{18} u_{2}]_{1}^{\frac{1}{e^{9}}}$

Step 4

Substitute and simplify.

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Step 4.1

Evaluate $- \frac{1}{18} u_{2}$ at $\frac{1}{e^{9}}$ and at $1$ .

$(- \frac{1}{18} \cdot \frac{1}{e^{9}}) + \frac{1}{18} \cdot 1$

Step 4.2

Simplify.

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Step 4.2.1

Multiply $\frac{1}{e^{9}}$ by $\frac{1}{18}$ .

$- \frac{1}{e^{9} \cdot 18} + \frac{1}{18} \cdot 1$

Step 4.2.2

Move $18$ to the left of $e^{9}$ .

$- \frac{1}{18 \cdot e^{9}} + \frac{1}{18} \cdot 1$

Step 4.2.3

Multiply $\frac{1}{18}$ by $1$ .

$- \frac{1}{18 e^{9}} + \frac{1}{18}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{18 e^{9}} + \frac{1}{18}$

Decimal Form:

$0.05554869 \dots$

Step 6