integral from -1 to 1 of 3x^2sqrt(x^3+5)

1) Let's compute the integral:

\int_{-1}^{1} 3 x^{2} \cdot \sqrt{x^{3} + 5} d x

Let

u = x^{3} + 5

. Then

du = 3 x^{2} \cdot dx

, so

3 x^{2} \cdot dx = du

\int 3 x^{2} \cdot \sqrt{x^{3} + 5} d x = \int \sqrt{u} d u

An antiderivative of

\sqrt{u}

\frac{2 u^{\frac{3}{2}}}{3}

Substitute

u

x^{3} + 5

back into the antiderivative

ⅰ

\int 3 x^{2} \cdot \sqrt{x^{3} + 5} d x

\frac{2 \cdot {(x^{3} + 5)}^{\frac{3}{2}}}{3}

▷

\frac{2 \cdot {(x^{3} + 5)}^{\frac{3}{2}}}{3}

2) Let's evaluate for the interval

[-1, 1]

\frac{2 \cdot {(1^{3} + 5)}^{\frac{3}{2}}}{3} - \frac{2 \cdot {({(-1)}^{3} + 5)}^{\frac{3}{2}}}{3}

ⓘ Parentheses are computed first:

1 + 5

6

■Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-1)}^{3}

(-1) \cdot (-1) \cdot (-1)

-1

\frac{2 \cdot 6^{\frac{3}{2}}}{3} - \frac{2 \cdot {(- 1 + 5)}^{\frac{3}{2}}}{3}

ⓘ Powers are computed before multiplications, additions and subtractions:

6^{\frac{3}{2}}

6 \cdot \sqrt{6}

■Parentheses are computed first:

- 1 + 5

4

\frac{2 \cdot 6 \cdot \sqrt{6}}{3} - \frac{2 \cdot 4^{\frac{3}{2}}}{3}

ⓘ Powers are computed before multiplications, additions and subtractions:

4^{\frac{3}{2}}

4 \cdot \sqrt{4}

ⅰ

2 \cdot 6

12

\frac{12 \cdot \sqrt{6}}{3} - \frac{2 \cdot 4 \cdot \sqrt{4}}{3}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{12 \cdot \sqrt{6}}{3}

4 \cdot \sqrt{6}

■The square root of a number is the positive number whose square is that number:

\sqrt{4}

2

;

2^{2}

4

ⅰ

4 \cdot 2

8

4 \cdot \sqrt{6} - \frac{8 \cdot 2}{3}

ⅰ

8 \cdot 2

16

▶Integral:

4 \cdot \sqrt{6} - \frac{16}{3}

3) The numeric value of:

4 \cdot \sqrt{6} - \frac{16}{3}

equals:

4 \cdot \approx 2.4495 - \approx 5.3333

\approx 9.7980 - \approx 5.3333

▷

\approx 4.4646