Calculus Examples

$\int (10^{20 t} - e^{20 t}) d t$

Step 1

Remove parentheses.

$\int 10^{20 t} - e^{20 t} d t$

Step 2

Split the single integral into multiple integrals.

$\int 10^{20 t} d t + \int - e^{20 t} d t$

Step 3

Let $u_{1} = 20 t$ . Then $d u_{1} = 20 d t$ , so $\frac{1}{20} d u_{1} = d t$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 3.1

Let $u_{1} = 20 t$ . Find $\frac{d u_{1}}{d t}$ .

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Step 3.1.1

Differentiate $20 t$ .

$\frac{d}{d t} [20 t]$

Step 3.1.2

Since $20$ is constant with respect to $t$ , the derivative of $20 t$ with respect to $t$ is $20 \frac{d}{d t} [t]$ .

$20 \frac{d}{d t} [t]$

Step 3.1.3

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

$20 \cdot 1$

Step 3.1.4

Multiply $20$ by $1$ .

$20$

Step 3.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int 10^{u_{1}} \cdot \frac{1}{20} d u_{1} + \int - e^{20 t} d t$

Step 4

Combine $10^{u_{1}}$ and $\frac{1}{20}$ .

$\int \frac{10^{u_{1}}}{20} d u_{1} + \int - e^{20 t} d t$

Step 5

Since $\frac{1}{20}$ is constant with respect to $u_{1}$ , move $\frac{1}{20}$ out of the integral.

$\frac{1}{20} \int 10^{u_{1}} d u_{1} + \int - e^{20 t} d t$

Step 6

The integral of $10^{u_{1}}$ with respect to $u_{1}$ is $\frac{10^{u_{1}}}{\ln (10)}$ .

$\frac{1}{20} (\frac{10^{u_{1}}}{\ln (10)} + C) + \int - e^{20 t} d t$

Step 7

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

$\frac{1}{20} (\frac{10^{u_{1}}}{\ln (10)} + C) - \int e^{20 t} d t$

Step 8

Let $u_{2} = 20 t$ . Then $d u_{2} = 20 d t$ , so $\frac{1}{20} d u_{2} = d t$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 8.1

Let $u_{2} = 20 t$ . Find $\frac{d u_{2}}{d t}$ .

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Step 8.1.1

Differentiate $20 t$ .

$\frac{d}{d t} [20 t]$

Step 8.1.2

Since $20$ is constant with respect to $t$ , the derivative of $20 t$ with respect to $t$ is $20 \frac{d}{d t} [t]$ .

$20 \frac{d}{d t} [t]$

Step 8.1.3

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

$20 \cdot 1$

Step 8.1.4

Multiply $20$ by $1$ .

$20$

Step 8.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{20} (\frac{10^{u_{1}}}{\ln (10)} + C) - \int e^{u_{2}} \frac{1}{20} d u_{2}$

Step 9

Combine $e^{u_{2}}$ and $\frac{1}{20}$ .

$\frac{1}{20} (\frac{10^{u_{1}}}{\ln (10)} + C) - \int \frac{e^{u_{2}}}{20} d u_{2}$

Step 10

Since $\frac{1}{20}$ is constant with respect to $u_{2}$ , move $\frac{1}{20}$ out of the integral.

$\frac{1}{20} (\frac{10^{u_{1}}}{\ln (10)} + C) - (\frac{1}{20} \int e^{u_{2}} d u_{2})$

Step 11

The integral of $e^{u_{2}}$ with respect to $u_{2}$ is $e^{u_{2}}$ .

$\frac{1}{20} (\frac{10^{u_{1}}}{\ln (10)} + C) - \frac{1}{20} (e^{u_{2}} + C)$

Step 12

Simplify.

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Step 12.1

Simplify.

$\frac{1 \cdot 10^{u_{1}}}{20 \ln (10)} - \frac{1}{20} e^{u_{2}} + C$

Step 12.2

Multiply $10^{u_{1}}$ by $1$ .

$\frac{10^{u_{1}}}{20 \ln (10)} - \frac{1}{20} e^{u_{2}} + C$

Step 13

Substitute back in for each integration substitution variable.

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Step 13.1

Replace all occurrences of $u_{1}$ with $20 t$ .

$\frac{10^{20 t}}{20 \ln (10)} - \frac{1}{20} e^{u_{2}} + C$

Step 13.2

Replace all occurrences of $u_{2}$ with $20 t$ .

$\frac{10^{20 t}}{20 \ln (10)} - \frac{1}{20} e^{20 t} + C$

Step 14

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

$\frac{10^{20 t}}{20 \ln (10)} - \frac{e^{20 t}}{20} + C$

Step 15

Reorder terms.

$\frac{1}{20 \ln (10)} \cdot 10^{20 t} - \frac{1}{20} e^{20 t} + C$