What is the integral from 0 to 1 of xsqrt(16x^2+9) ?

The integral from 0 to 1 of xsqrt(16x^2+9) is 49/24

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integral from 0 to 1 of xsqrt(16x^2+9)

\int_{0}^{1} x 16 x^{2} + 9 d x

\frac{49}{24}

Decimal

2.04166 \dots

Solution steps

\int_{0}^{1} x 16 x^{2} + 9 d x

Apply u-substitution

= \int_{9}^{25} \frac{u}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int_{9}^{25} u d u

Apply the Power Rule

= \frac{1}{32} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

Compute the boundaries:

\frac{196}{3}

= \frac{1}{32} \cdot \frac{196}{3}

Simplify

= \frac{49}{24}

What is the integral from 0 to 1 of xsqrt(16x^2+9) ?
The integral from 0 to 1 of xsqrt(16x^2+9) is 49/24