integral from -8 to 0 of (y/8+sqrt(y+9))

1) Let's integrate:

\int_{-8}^{0} (\frac{y}{8} + \sqrt{y + 9}) d y

For the math expression:

\sqrt{y + 9}

, since (a^n)' = n*a'*a^(n - 1), we try differentiating

{(y + 9)}^{\frac{1}{2} + 1}

2) Let's perform the differentiation:

\frac{d}{d y} ({(y + 9)}^{\frac{1}{2} + 1})

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1}{2} + 1

\frac{1 + 2}{2}

\frac{d}{d y} ({(y + 9)}^{\frac{1 + 2}{2}})

ⓘ The numerator and the denominator are computed before the division:

1 + 2

3

\frac{d}{d y} ({(y + 9)}^{\frac{3}{2}})

Apply the differentiation rule:

\frac{d}{d y} (f^{n}) = n \cdot \frac{d}{d y} (f) \cdot f^{n - 1}

with f =

y + 9

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d y} ({(y + 9)}^{\frac{3}{2}})

\frac{3}{2} \cdot \frac{d}{d y} (y + 9) \cdot \sqrt{y + 9}

\frac{3}{2} \cdot \frac{d}{d y} (y + 9) \cdot \sqrt{y + 9}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{3}{2} \cdot \frac{d}{d y} (y + 9) \cdot \sqrt{y + 9}

\frac{3 \cdot \frac{d}{d y} (y + 9) \cdot \sqrt{y + 9}}{2}

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d y} (y + 9)

1 + 0

\frac{3 \cdot (1 + 0) \cdot \sqrt{y + 9}}{2}

ⓘ Adding 0 changes nothing:

1 + 0

1

\frac{3 \cdot \sqrt{y + 9}}{2}

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

ⅰ

\int (\frac{y}{8} + \sqrt{y + 9}) d y

\frac{2}{3} \cdot {(y + 9)}^{\frac{3}{2}} + \int \frac{y}{8} d y

\frac{2}{3} \cdot {(y + 9)}^{\frac{3}{2}} + \int \frac{y}{8} d y

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{2}{3} \cdot {(y + 9)}^{\frac{3}{2}}

\frac{2 \cdot {(y + 9)}^{\frac{3}{2}}}{3}

■A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int \frac{y}{8} d y

\frac{1}{8} \cdot \frac{y^{2}}{2}

\frac{2 \cdot {(y + 9)}^{\frac{3}{2}}}{3} + \frac{1}{8} \cdot \frac{y^{2}}{2}

ⓘ To multiply fractions, multiply the numerators together and the denominators together:

\frac{1}{8} \cdot \frac{y^{2}}{2}

\frac{1 \cdot y^{2}}{8 \cdot 2}

\frac{y^{2}}{16}

▷

\frac{2 \cdot {(y + 9)}^{\frac{3}{2}}}{3} + \frac{y^{2}}{16}

3) Let's evaluate over the interval

[-8, 0]

(\frac{0^{2}}{16} + \frac{2 \cdot {(0 + 9)}^{\frac{3}{2}}}{3}) - (\frac{{(-8)}^{2}}{16} + \frac{2 \cdot {(- 8 + 9)}^{\frac{3}{2}}}{3})

ⓘ Adding 0 changes nothing:

0 + 9

9

■Powers are computed before multiplications, additions and subtractions:

9^{\frac{3}{2}}

9 \cdot \sqrt{9}

■A minus sign before a parenthesis changes the sign of each of its terms:

- (\frac{{(-8)}^{2}}{16} + \frac{2 \cdot {(- 8 + 9)}^{\frac{3}{2}}}{3})

- \frac{{(-8)}^{2}}{16} - \frac{2 \cdot {(- 8 + 9)}^{\frac{3}{2}}}{3}

0

+ \frac{2 \cdot 9 \cdot \sqrt{9}}{3}

- \frac{{(-8)}^{2}}{16}

- \frac{2 \cdot {(- 8 + 9)}^{\frac{3}{2}}}{3}

ⓘ Adding 0 changes nothing:

0 + \frac{2 \cdot 9 \cdot \sqrt{9}}{3}

\frac{2 \cdot 9 \cdot \sqrt{9}}{3}

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

\sqrt{9}

3

;

3^{2}

9

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

{(-8)}^{2}

(-8) \cdot (-8)

64

■Parentheses are computed first:

- 8 + 9

1

ⅰ

2 \cdot 9

18

\frac{18 \cdot 3}{3} - \frac{64}{16} - \frac{2 \cdot 1^{\frac{3}{2}}}{3}

We simplify the fraction:

\frac{64}{16}

by the common factor:

16

to get the new expression:

4

\frac{54}{3} - 4 - \frac{2}{3}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

3

, their greatest common divisor:

\frac{54}{3}

18

18 - 4 - \frac{2}{3}

ⓘ Start from

18

and move back

4

on the number line:

18 - 4

14

14 - \frac{2}{3}

ⓘ To add fractions, put them over the same denominator, then add the numerators:

14 - \frac{2}{3}

\frac{3 \cdot 14 - 2}{3}

\frac{3 \cdot 14 - 2}{3}

ⓘ Multiplications are done before additions and subtractions:

3 \cdot 14

3 \cdot (10 + 4)

3 \cdot 10 + 3 \cdot 4

30 + 12

42

\frac{42 - 2}{3}

ⓘ The numerator and the denominator are computed before the division:

42 - 2

40

▶Integral:

\frac{40}{3}

4) The numeric value of:

\frac{40}{3}

is:

▷

\approx 13.3333