What is the integral from 0 to 1 of xsqrt(40x^2+9) ?

The integral from 0 to 1 of xsqrt(40x^2+9) is 79/30

Popular >

integral from 0 to 1 of xsqrt(40x^2+9)

\int_{0}^{1} x 40 x^{2} + 9 d x

\frac{79}{30}

Decimal

2.63333 \dots

Solution steps

\int_{0}^{1} x 40 x^{2} + 9 d x

Apply u-substitution

: \int_{9}^{49} \frac{u}{80} d u

= \int_{9}^{49} \frac{u}{80} d u

Take the constant out:

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

= \frac{1}{80} \cdot \int_{9}^{49} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{49}

= \frac{1}{80} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{49}

Compute the boundaries:

\frac{632}{3}

= \frac{1}{80} \cdot \frac{632}{3}

Simplify

= \frac{79}{30}

\int_{7}^{25} x \frac{x - 7}{162} d x

\int_{0}^{4} x^{3} 256 - x^{4} d x

\int_{1}^{e} \frac{3 ln ( x ^{2} )}{x} d x

\int_{0}^{0.4} 1 + x^{2} d x

\int_{- 2 π}^{4 π} cos^{2} (x) d x

What is the integral from 0 to 1 of xsqrt(40x^2+9) ?
The integral from 0 to 1 of xsqrt(40x^2+9) is 79/30