Calculus Examples

$\int_{0}^{1} v^{2} \cos (v^{3}) d v$

Step 1

Let $u = v^{3}$ . Then $d u = 3 v^{2} d v$ , so $\frac{1}{3} d u = v^{2} d v$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = v^{3}$ . Find $\frac{d u}{d v}$ .

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Step 1.1.1

Differentiate $v^{3}$ .

$\frac{d}{d v} [v^{3}]$

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d v} [v^{n}]$ is $n v^{n - 1}$ where $n = 3$ .

$3 v^{2}$

Step 1.2

Substitute the lower limit in for $v$ in $u = v^{3}$ .

$u_{lower} = 0^{3}$

Step 1.3

Raising $0$ to any positive power yields $0$ .

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $v$ in $u = v^{3}$ .

$u_{upper} = 1^{3}$

Step 1.5

One to any power is one.

$u_{upper} = 1$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = 1$

Step 1.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{0}^{1} \cos (u) \frac{1}{3} d u$

Step 2

Combine $\cos (u)$ and $\frac{1}{3}$ .

$\int_{0}^{1} \frac{\cos (u)}{3} d u$

Step 3

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int_{0}^{1} \cos (u) d u$

Step 4

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$\frac{1}{3} \sin (u)]_{0}^{1}$

Step 5

Evaluate $\sin (u)$ at $1$ and at $0$ .

$\frac{1}{3} (\sin (1) - \sin (0))$

Step 6

Simplify.

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Step 6.1

The exact value of $\sin (0)$ is $0$ .

$\frac{1}{3} (\sin (1) - 0)$

Step 6.2

Multiply $- 1$ by $0$ .

$\frac{1}{3} (\sin (1) + 0)$

Step 6.3

Add $\sin (1)$ and $0$ .

$\frac{1}{3} \sin (1)$

Step 6.4

Combine $\frac{1}{3}$ and $\sin (1)$ .

$\frac{\sin (1)}{3}$

Step 7

Simplify.

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Step 7.1

Evaluate $\sin (1)$ .

$\frac{0.84147098}{3}$

Step 7.2

Divide $0.84147098$ by $3$ .

$0.28049032$