Calculus Examples

$a' (t) = (\frac{15}{2}) \sqrt{t} + 3 e^{- t}$

Step 1

The function $a (t)$ can be found by evaluating the indefinite integral of the derivative $a' (t)$ .

$a (t) = \int a' (t) d t$

Step 2

Combine $\frac{15}{2}$ and $\sqrt{t}$ .

$\int \frac{15 \sqrt{t}}{2} + 3 e^{- t} d t$

Step 3

Split the single integral into multiple integrals.

$\int \frac{15 \sqrt{t}}{2} d t + \int 3 e^{- t} d t$

Step 4

Since $\frac{15}{2}$ is constant with respect to $t$ , move $\frac{15}{2}$ out of the integral.

$\frac{15}{2} \int \sqrt{t} d t + \int 3 e^{- t} d t$

Step 5

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t}$ as $t^{\frac{1}{2}}$ .

$\frac{15}{2} \int t^{\frac{1}{2}} d t + \int 3 e^{- t} d t$

Step 6

By the Power Rule, the integral of $t^{\frac{1}{2}}$ with respect to $t$ is $\frac{2}{3} t^{\frac{3}{2}}$ .

$\frac{15}{2} (\frac{2}{3} t^{\frac{3}{2}} + C) + \int 3 e^{- t} d t$

Step 7

Since $3$ is constant with respect to $t$ , move $3$ out of the integral.

$\frac{15}{2} (\frac{2}{3} t^{\frac{3}{2}} + C) + 3 \int e^{- t} d t$

Step 8

Let $u = - t$ . Then $d u = - d t$ , so $- d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

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

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Step 8.1.1

Differentiate $- t$ .

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

Step 8.1.2

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

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

Step 8.1.3

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

$- 1 \cdot 1$

Step 8.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 8.2

Rewrite the problem using $u$ and $d u$ .

$\frac{15}{2} (\frac{2}{3} t^{\frac{3}{2}} + C) + 3 \int - e^{u} d u$

Step 9

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

$\frac{15}{2} (\frac{2}{3} t^{\frac{3}{2}} + C) + 3 (- \int e^{u} d u)$

Step 10

Multiply $- 1$ by $3$ .

$\frac{15}{2} (\frac{2}{3} t^{\frac{3}{2}} + C) - 3 \int e^{u} d u$

Step 11

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

$\frac{15}{2} (\frac{2}{3} t^{\frac{3}{2}} + C) - 3 (e^{u} + C)$

Step 12

Simplify.

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Step 12.1

Simplify.

$\frac{15}{2} \cdot \frac{2}{3} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2

Simplify.

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Step 12.2.1

Multiply $\frac{15}{2}$ by $\frac{2}{3}$ .

$\frac{15 \cdot 2}{2 \cdot 3} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2.2

Multiply $15$ by $2$ .

$\frac{30}{2 \cdot 3} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2.3

Multiply $2$ by $3$ .

$\frac{30}{6} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2.4

Cancel the common factor of $30$ and $6$ .

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Step 12.2.4.1

Factor $6$ out of $30$ .

$\frac{6 \cdot 5}{6} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2.4.2

Cancel the common factors.

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Step 12.2.4.2.1

Factor $6$ out of $6$ .

$\frac{6 \cdot 5}{6 (1)} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2.4.2.2

Cancel the common factor.

$\frac{6 \cdot 5}{6 \cdot 1} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2.4.2.3

Rewrite the expression.

$\frac{5}{1} t^{\frac{3}{2}} - 3 e^{u} + C$

Step 12.2.4.2.4

Divide $5$ by $1$ .

$5 t^{\frac{3}{2}} - 3 e^{u} + C$

Step 13

Replace all occurrences of $u$ with $- t$ .

$5 t^{\frac{3}{2}} - 3 e^{- t} + C$

Step 14

The function $a$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$a (t) = 5 t^{\frac{3}{2}} - 3 e^{- t} + C$