integral from 0 to 1 of sqrt(x+16) — step-by-step solution

1) Let's integrate:

\int_{0}^{1} \sqrt{x + 16} d x

For the math expression:

\sqrt{x + 16}

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

{(x + 16)}^{\frac{1}{2} + 1}

2) Let's differentiate:

\frac{d}{d x} ({(x + 16)}^{\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 x} ({(x + 16)}^{\frac{1 + 2}{2}})

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

1 + 2

3

\frac{d}{d x} ({(x + 16)}^{\frac{3}{2}})

We differentiate using:

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

(where f =

x + 16

)

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d x} ({(x + 16)}^{\frac{3}{2}})

\frac{3}{2} \cdot \frac{d}{d x} (x + 16) \cdot \sqrt{x + 16}

\frac{3}{2} \cdot \frac{d}{d x} (x + 16) \cdot \sqrt{x + 16}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{3}{2} \cdot \frac{d}{d x} (x + 16) \cdot \sqrt{x + 16}

\frac{3 \cdot \frac{d}{d x} (x + 16) \cdot \sqrt{x + 16}}{2}

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

\frac{d}{d x} (x + 16)

1 + 0

\frac{3 \cdot (1 + 0) \cdot \sqrt{x + 16}}{2}

ⓘ Adding 0 changes nothing:

1 + 0

1

\frac{3 \cdot \sqrt{x + 16}}{2}

The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:

ⓘ The derivative of

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

\sqrt{x + 16}

, so

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

is an antiderivative of

\sqrt{x + 16}

\int \sqrt{x + 16} d x

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

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

ⓘ Multiplying a fraction by a term multiplies its numerator:

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

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

▷

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

3) Let's evaluate over the interval

[0, 1]

\frac{2 \cdot {(1 + 16)}^{\frac{3}{2}}}{3} - \frac{2 \cdot {(0 + 16)}^{\frac{3}{2}}}{3}

ⓘ Parentheses are computed first:

1 + 16

17

■Adding 0 changes nothing:

0 + 16

16

■Powers are computed before multiplications, additions and subtractions:

16^{\frac{3}{2}}

16 \cdot \sqrt{16}

\frac{2 \cdot 17^{\frac{3}{2}}}{3} - \frac{2 \cdot 16 \cdot \sqrt{16}}{3}

ⓘ Powers are computed before multiplications, additions and subtractions:

17^{\frac{3}{2}}

17 \cdot \sqrt{17}

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

\sqrt{16}

4

;

4^{2}

16

ⅰ

2 \cdot 16

2 \cdot (10 + 6)

2 \cdot 10 + 2 \cdot 6

20 + 12

32

\frac{2 \cdot 17 \cdot \sqrt{17}}{3} - \frac{32 \cdot 4}{3}

ⅰ

2 \cdot 17

2 \cdot (10 + 7)

2 \cdot 10 + 2 \cdot 7

20 + 14

34

■

32 \cdot 4

(30 + 2) \cdot 4

30 \cdot 4 + 2 \cdot 4

120 + 8

128

\frac{34 \cdot \sqrt{17}}{3} - \frac{128}{3}

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

\frac{34 \cdot \sqrt{17}}{3} - \frac{128}{3}

\frac{34 \cdot \sqrt{17} - 128}{3}

▶Integral:

\frac{34 \cdot \sqrt{17} - 128}{3}

4) The numeric value of:

\frac{34 \cdot \sqrt{17} - 128}{3}

is:

\frac{34 \cdot \approx 4.1231 - 128}{3}

\frac{\approx 140.1856 - 128}{3}

\frac{\approx 12.1856}{3}

▷

\approx 4.0619