Calculus Examples

$p (t) = t^{3} - \frac{t^{2}}{2} - t$

Step 1

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

$P (t) = \int p (t) d t$

Step 2

Set up the integral to solve.

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

Step 3

Split the single integral into multiple integrals.

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

Step 4

By the Power Rule, the integral of $t^{3}$ with respect to $t$ is $\frac{1}{4} t^{4}$ .

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

Step 5

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

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

Step 6

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

$\frac{1}{4} t^{4} + C - (\frac{1}{2} \int t^{2} d t) + \int - t d t$

Step 7

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

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

Step 8

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

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

Step 9

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

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

Step 10

Simplify.

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Step 10.1

Simplify.

$\frac{t^{4}}{4} - \frac{t^{3}}{6} - (\frac{1}{2} t^{2}) + C$

Step 10.2

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

$\frac{t^{4}}{4} - \frac{t^{3}}{6} - \frac{t^{2}}{2} + C$

Step 11

Reorder terms.

$\frac{1}{4} t^{4} - \frac{1}{6} t^{3} - \frac{1}{2} t^{2} + C$

Step 12

The answer is the antiderivative of the function $p (t) = t^{3} - \frac{t^{2}}{2} - t$ .

$P (t) =$ $\frac{1}{4} t^{4} - \frac{1}{6} t^{3} - \frac{1}{2} t^{2} + C$