What is the integral from 0 to 1 of xsqrt(8-x^2) ?

The integral from 0 to 1 of xsqrt(8-x^2) is (16sqrt(2)-7sqrt(7))/3

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integral from 0 to 1 of xsqrt(8-x^2)

\int_{0}^{1} x 8 - x^{2} d x

\frac{16 2 - 7 7}{3}

Solution steps

\int_{0}^{1} x 8 - x^{2} d x

Apply u-substitution

: \int_{8}^{7} - \frac{u}{2} d u

= \int_{8}^{7} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{7}^{8} - \frac{u}{2} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{7}^{8} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{8}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{8})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{8}

Compute the boundaries:

\frac{32 2 - 14 7}{3}

= \frac{1}{2} \cdot \frac{32 2 - 14 7}{3}

Simplify

= \frac{16 2 - 7 7}{3}

(1 + ln (x) + \frac{y}{x}) d x = (7 - ln (x)) d y

x \to - 7 lim (\frac{1}{x + 7})

[t]

\frac{d}{d x} (ln (e^{6 x}))

\int (\frac{1}{9 - x ^{2}}) d x

What is the integral from 0 to 1 of xsqrt(8-x^2) ?
The integral from 0 to 1 of xsqrt(8-x^2) is (16sqrt(2)-7sqrt(7))/3