What is the integral from 0 to 3/4 of sqrt(1+(2x^{1/2))^2} ?

The integral from 0 to 3/4 of sqrt(1+(2x^{1/2))^2} is 7/6

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integral from 0 to 3/4 of sqrt(1+(2x^{1/2))^2}

\int_{0}^{\frac{3}{4}} 1 + (2 x^{\frac{1}{2}})^{2} d x

\frac{7}{6}

Decimal

1.16666 \dots

Solution steps

\int_{0}^{\frac{3}{4}} 1 + (2 x^{\frac{1}{2}})^{2} d x

Simplify

1 + (2 x^{\frac{1}{2}})^{2} : 1 + 4 x

= \int_{0}^{\frac{3}{4}} 1 + 4 x d x

Apply u-substitution

: \int_{1}^{4} \frac{u}{4} d u

= \int_{1}^{4} \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{1}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{1}{4} \cdot \frac{14}{3}

Simplify

= \frac{7}{6}

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What is the integral from 0 to 3/4 of sqrt(1+(2x^{1/2))^2} ?
The integral from 0 to 3/4 of sqrt(1+(2x^{1/2))^2} is 7/6