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

The integral from 0 to 1 of sqrt(8x+1) is 13/6

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

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

\frac{13}{6}

Solution steps

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

Apply u-substitution

: \int_{1}^{9} \frac{u}{8} d u

= \int_{1}^{9} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{1}{8} \cdot \frac{52}{3}

Simplify

= \frac{13}{6}

\int_{- 2}^{1} x - 8 ∣ x ∣ d x

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\int_{1}^{2} x^{2} - 2 x d x

\int_{2}^{9} (4 x + 10) d x

What is the integral from 0 to 1 of sqrt(8x+1) ?
The integral from 0 to 1 of sqrt(8x+1) is 13/6