What is the integral from 0 to 8 of (12)/(sqrt(x+1)) ?

The integral from 0 to 8 of (12)/(sqrt(x+1)) is 48

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integral from 0 to 8 of (12)/(sqrt(x+1))

\int_{0}^{8} \frac{12}{x + 1} d x

48

Solution steps

\int_{0}^{8} \frac{12}{x + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int_{0}^{8} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{1}^{3} 2 d u

= 12 \cdot \int_{1}^{3} 2 d u

Integral of a constant:

\int a d x = a x

= 12 [2 u]_{1}^{3}

Compute the boundaries:

4

= 12 \cdot 4

Simplify

= 48

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What is the integral from 0 to 8 of (12)/(sqrt(x+1)) ?
The integral from 0 to 8 of (12)/(sqrt(x+1)) is 48