What is the integral from 0 to 15/16 of sqrt(1+16x) ?

The integral from 0 to 15/16 of sqrt(1+16x) is 21/8

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integral from 0 to 15/16 of sqrt(1+16x)

\int_{0}^{\frac{15}{16}} 1 + 16 x d x

\frac{21}{8}

Decimal

2.625

Solution steps

\int_{0}^{\frac{15}{16}} 1 + 16 x d x

Apply u-substitution

= \int_{1}^{16} \frac{u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{1}^{16} u d u

Apply the Power Rule

= \frac{1}{16} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{16}

Compute the boundaries:

42

= \frac{1}{16} \cdot 42

Simplify

= \frac{21}{8}

What is the integral from 0 to 15/16 of sqrt(1+16x) ?
The integral from 0 to 15/16 of sqrt(1+16x) is 21/8