What is the integral from 0 to 1 of xsqrt(45x^2+36) ?

The integral from 0 to 1 of xsqrt(45x^2+36) is 19/5

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integral from 0 to 1 of xsqrt(45x^2+36)

\int_{0}^{1} x 45 x^{2} + 36 d x

\frac{19}{5}

Decimal

3.8

Solution steps

\int_{0}^{1} x 45 x^{2} + 36 d x

Apply u-substitution

= \int_{36}^{81} \frac{u}{90} d u

Take the constant out:

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

= \frac{1}{90} \cdot \int_{36}^{81} u d u

Apply the Power Rule

= \frac{1}{90} [\frac{2}{3} u^{\frac{3}{2}}]_{36}^{81}

Compute the boundaries:

342

= \frac{1}{90} \cdot 342

Simplify

= \frac{19}{5}

What is the integral from 0 to 1 of xsqrt(45x^2+36) ?
The integral from 0 to 1 of xsqrt(45x^2+36) is 19/5