Calculus Examples

$(x - 1) \sqrt{x}$

Step 1

Write $(x - 1) \sqrt{x}$ as a function.

$f (x) = (x - 1) \sqrt{x}$

Step 2

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

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

Step 3

Set up the integral to solve.

$F (x) = \int (x - 1) \sqrt{x} d x$

Step 4

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

$\int (x - 1) x^{\frac{1}{2}} d x$

Step 5

Expand $(x - 1) x^{\frac{1}{2}}$ .

Tap for more steps...

Step 5.1

Apply the distributive property.

$\int x \cdot x^{\frac{1}{2}} - 1 x^{\frac{1}{2}} d x$

Step 5.2

Raise $x$ to the power of $1$ .

$\int x^{1} x^{\frac{1}{2}} - 1 x^{\frac{1}{2}} d x$

Step 5.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int x^{1 + \frac{1}{2}} - 1 x^{\frac{1}{2}} d x$

Step 5.4

Write $1$ as a fraction with a common denominator.

$\int x^{\frac{2}{2} + \frac{1}{2}} - 1 x^{\frac{1}{2}} d x$

Step 5.5

Combine the numerators over the common denominator.

$\int x^{\frac{2 + 1}{2}} - 1 x^{\frac{1}{2}} d x$

Step 5.6

Add $2$ and $1$ .

$\int x^{\frac{3}{2}} - 1 x^{\frac{1}{2}} d x$

Step 6

Split the single integral into multiple integrals.

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

Step 7

By the Power Rule, the integral of $x^{\frac{3}{2}}$ with respect to $x$ is $\frac{2}{5} x^{\frac{5}{2}}$ .

$\frac{2}{5} x^{\frac{5}{2}} + C + \int - 1 x^{\frac{1}{2}} d x$

Step 8

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

$\frac{2}{5} x^{\frac{5}{2}} + C - \int x^{\frac{1}{2}} d x$

Step 9

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

$\frac{2}{5} x^{\frac{5}{2}} + C - (\frac{2}{3} x^{\frac{3}{2}} + C)$

Step 10

Simplify.

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

Step 11

The answer is the antiderivative of the function $f (x) = (x - 1) \sqrt{x}$ .

$F (x) =$ $\frac{2}{5} x^{\frac{5}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C$