Calculus Examples

$\frac{y^{3}}{2} + \frac{1}{2 y^{3}}$

Step 1

Write $\frac{y^{3}}{2} + \frac{1}{2 y^{3}}$ as a function.

$f (y) = \frac{y^{3}}{2} + \frac{1}{2 y^{3}}$

Step 2

The function $F (y)$ can be found by finding the indefinite integral of the derivative $f (y)$ .

$F (y) = \int f (y) d y$

Step 3

Set up the integral to solve.

$F (y) = \int \frac{y^{3}}{2} + \frac{1}{2 y^{3}} d y$

Step 4

Split the single integral into multiple integrals.

$\int \frac{y^{3}}{2} d y + \int \frac{1}{2 y^{3}} d y$

Step 5

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

$\frac{1}{2} \int y^{3} d y + \int \frac{1}{2 y^{3}} d y$

Step 6

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

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

Step 7

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

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

Step 8

Apply basic rules of exponents.

Tap for more steps...

Step 8.1

Move $y^{3}$ out of the denominator by raising it to the $- 1$ power.

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

Step 8.2

Multiply the exponents in ${(y^{3})}^{- 1}$ .

Tap for more steps...

Step 8.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 8.2.2

Multiply $3$ by $- 1$ .

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

Step 9

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

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

Step 10

Simplify.

Tap for more steps...

Step 10.1

Simplify.

$\frac{1}{2} \cdot \frac{1}{4} y^{4} + \frac{1}{2} (- \frac{1}{2} y^{- 2}) + C$

Step 10.2

Simplify.

Tap for more steps...

Step 10.2.1

Multiply $\frac{1}{2}$ by $\frac{1}{4}$ .

$\frac{1}{2 \cdot 4} y^{4} + \frac{1}{2} (- \frac{1}{2} y^{- 2}) + C$

Step 10.2.2

Multiply $2$ by $4$ .

$\frac{1}{8} y^{4} + \frac{1}{2} (- \frac{1}{2} y^{- 2}) + C$

Step 10.2.3

Combine $y^{- 2}$ and $\frac{1}{2}$ .

$\frac{1}{8} y^{4} + \frac{1}{2} (- \frac{y^{- 2}}{2}) + C$

Step 10.2.4

Multiply $\frac{1}{2}$ by $\frac{y^{- 2}}{2}$ .

$\frac{1}{8} y^{4} - \frac{y^{- 2}}{2 \cdot 2} + C$

Step 10.2.5

Multiply $2$ by $2$ .

$\frac{1}{8} y^{4} - \frac{y^{- 2}}{4} + C$

Step 10.2.6

Move $y^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{8} y^{4} - \frac{1}{4 y^{2}} + C$

Step 11

The answer is the antiderivative of the function $f (y) = \frac{y^{3}}{2} + \frac{1}{2 y^{3}}$ .

$F (y) =$ $\frac{1}{8} y^{4} - \frac{1}{4 y^{2}} + C$