Calculus Examples

$S (t) = t^{2} - \frac{t}{2} + \sqrt{t}$

Step 1

The function $3 (t)$ can be found by finding the indefinite integral of the derivative $S (t)$ .

$3 (t) = \int S (t) d t$

Step 2

Set up the integral to solve.

$3 (t) = \int t^{2} - \frac{t}{2} + \sqrt{t} d t$

Step 3

Split the single integral into multiple integrals.

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

Step 4

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

$\frac{1}{3} t^{3} + C + \int - \frac{t}{2} d t + \int \sqrt{t} d t$

Step 5

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

$\frac{1}{3} t^{3} + C - \int \frac{t}{2} d t + \int \sqrt{t} d t$

Step 6

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

$\frac{1}{3} t^{3} + C - (\frac{1}{2} \int t d t) + \int \sqrt{t} d t$

Step 7

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

$\frac{1}{3} t^{3} + C - \frac{1}{2} (\frac{1}{2} t^{2} + C) + \int \sqrt{t} d t$

Step 8

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

$\frac{1}{3} t^{3} + C - \frac{1}{2} (\frac{1}{2} t^{2} + C) + \int t^{\frac{1}{2}} d t$

Step 9

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

$\frac{1}{3} t^{3} + C - \frac{1}{2} (\frac{1}{2} t^{2} + C) + \frac{2}{3} t^{\frac{3}{2}} + C$

Step 10

Simplify.

$\frac{t^{3}}{3} - \frac{t^{2}}{4} + \frac{2}{3} t^{\frac{3}{2}} + C$

Step 11

Reorder terms.

$\frac{1}{3} t^{3} - \frac{1}{4} t^{2} + \frac{2}{3} t^{\frac{3}{2}} + C$

Step 12

The answer is the antiderivative of the function $S (t) = t^{2} - \frac{t}{2} + \sqrt{t}$ .

$3 (t) =$ $\frac{1}{3} t^{3} - \frac{1}{4} t^{2} + \frac{2}{3} t^{\frac{3}{2}} + C$