Calculus Examples

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

Step 1

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

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Step 1.1

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

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Step 1.1.1

Differentiate $1 - 2 x$ .

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

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 $n = 1$ .

$0 - 2 \cdot 1$

Step 1.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 1.1.4

Subtract $2$ from $0$ .

$- 2$

Step 1.2

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

$u_{lower} = 1 - 2 \cdot 0$

Step 1.3

Simplify.

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Step 1.3.1

Multiply $- 2$ by $0$ .

$u_{lower} = 1 + 0$

Step 1.3.2

Add $1$ and $0$ .

$u_{lower} = 1$

Step 1.4

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

$u_{upper} = 1 - 2 \cdot 1$

Step 1.5

Simplify.

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Step 1.5.1

Multiply $- 2$ by $1$ .

$u_{upper} = 1 - 2$

Step 1.5.2

Subtract $2$ from $1$ .

$u_{upper} = - 1$

Step 1.6

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

$u_{lower} = 1$

$u_{upper} = - 1$

Step 1.7

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

$\int_{1}^{- 1} e^{u} \frac{1}{- 2} d u$

Step 2

Simplify.

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Step 2.1

Move the negative in front of the fraction.

$\int_{1}^{- 1} e^{u} (- \frac{1}{2}) d u$

Step 2.2

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

$\int_{1}^{- 1} - \frac{e^{u}}{2} d u$

Step 3

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

$- \int_{1}^{- 1} \frac{e^{u}}{2} d u$

Step 4

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

$- (\frac{1}{2} \int_{1}^{- 1} e^{u} d u)$

Step 5

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

$- \frac{1}{2} e^{u}]_{1}^{- 1}$

Step 6

Substitute and simplify.

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Step 6.1

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

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

Step 6.2

Simplify.

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

Step 7

Simplify.

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Step 7.1

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

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

Step 7.2

Apply the distributive property.

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

Step 7.3

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

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

Step 7.4

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

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Step 7.4.1

Multiply $- 1$ by $- 1$ .

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

Step 7.4.2

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

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

Step 7.4.3

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

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

Step 7.5

Move $2$ to the left of $e$ .

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

Step 8

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$1.17520119 \dots$

Step 9