Calculus Examples

$\int_{0}^{6} 10 e^{- \frac{1}{2} t} d t$

Step 1

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

$\int_{0}^{6} 10 e^{- \frac{t}{2}} d t$

Step 2

Since $10$ is constant with respect to $t$ , move $10$ out of the integral.

$10 \int_{0}^{6} e^{- \frac{t}{2}} d t$

Step 3

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

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Step 3.1

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

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Step 3.1.1

Differentiate $- \frac{t}{2}$ .

$\frac{d}{d t} [- \frac{t}{2}]$

Step 3.1.2

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

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

Step 3.1.3

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

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

Step 3.1.4

Multiply $- 1$ by $1$ .

$- \frac{1}{2}$

Step 3.2

Substitute the lower limit in for $t$ in $u = - \frac{t}{2}$ .

$u_{lower} = - \frac{0}{2}$

Step 3.3

Simplify.

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Step 3.3.1

Divide $0$ by $2$ .

$u_{lower} = - 0$

Step 3.3.2

Multiply $- 1$ by $0$ .

$u_{lower} = 0$

Step 3.4

Substitute the upper limit in for $t$ in $u = - \frac{t}{2}$ .

$u_{upper} = - \frac{6}{2}$

Step 3.5

Simplify.

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Step 3.5.1

Divide $6$ by $2$ .

$u_{upper} = - 1 \cdot 3$

Step 3.5.2

Multiply $- 1$ by $3$ .

$u_{upper} = - 3$

Step 3.6

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

$u_{lower} = 0$

$u_{upper} = - 3$

Step 3.7

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

$10 \int_{0}^{- 3} - e^{u} \frac{- 1}{- \frac{1}{2}} d u$

Step 4

Simplify.

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Step 4.1

Dividing two negative values results in a positive value.

$10 \int_{0}^{- 3} - e^{u} \frac{1}{\frac{1}{2}} d u$

Step 4.2

Multiply by the reciprocal of the fraction to divide by $\frac{1}{2}$ .

$10 \int_{0}^{- 3} - e^{u} (1 \cdot 2) d u$

Step 4.3

Multiply $2$ by $1$ .

$10 \int_{0}^{- 3} - e^{u} \cdot 2 d u$

Step 4.4

Multiply $2$ by $- 1$ .

$10 \int_{0}^{- 3} - 2 e^{u} d u$

Step 5

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

$10 (- 2 \int_{0}^{- 3} e^{u} d u)$

Step 6

Multiply $- 2$ by $10$ .

$- 20 \int_{0}^{- 3} e^{u} d u$

Step 7

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

$- 20 (e^{u}]_{0}^{- 3})$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $e^{u}$ at $- 3$ and at $0$ .

$- 20 ((e^{- 3}) - e^{0})$

Step 8.2

Simplify.

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Step 8.2.1

Anything raised to $0$ is $1$ .

$- 20 (e^{- 3} - 1 \cdot 1)$

Step 8.2.2

Multiply $- 1$ by $1$ .

$- 20 (e^{- 3} - 1)$

Step 9

Simplify.

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Step 9.1

Apply the distributive property.

$- 20 e^{- 3} - 20 \cdot - 1$

Step 9.2

Multiply $- 20$ by $- 1$ .

$- 20 e^{- 3} + 20$

Step 9.3

Simplify each term.

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Step 9.3.1

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

$- 20 \frac{1}{e^{3}} + 20$

Step 9.3.2

Combine $- 20$ and $\frac{1}{e^{3}}$ .

$\frac{- 20}{e^{3}} + 20$

Step 9.3.3

Move the negative in front of the fraction.

$- \frac{20}{e^{3}} + 20$

Step 10

The result can be shown in multiple forms.

Exact Form:

$- \frac{20}{e^{3}} + 20$

Decimal Form:

$19.00425863 \dots$

Step 11