What is the integral from 1 to 2 of 5xsqrt(5-x^2) ?

The integral from 1 to 2 of 5xsqrt(5-x^2) is 35/3

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integral from 1 to 2 of 5xsqrt(5-x^2)

\int_{1}^{2} 5 x 5 - x^{2} d x

\frac{35}{3}

Decimal

11.66666 \dots

Solution steps

\int_{1}^{2} 5 x 5 - x^{2} d x

Take the constant out:

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

= 5 \cdot \int_{1}^{2} x 5 - x^{2} d x

Apply u-substitution

: \int_{4}^{1} - \frac{u}{2} d u

= 5 \cdot \int_{4}^{1} - \frac{u}{2} d u

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

= 5 (- \int_{1}^{4} - \frac{u}{2} d u)

Take the constant out:

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

= 5 (- (- \frac{1}{2} \cdot \int_{1}^{4} u d u))

Apply the Power Rule

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

= 5 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}))

Simplify

5 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4})) : \frac{5}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= \frac{5}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{5}{2} \cdot \frac{14}{3}

Simplify

= \frac{35}{3}

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What is the integral from 1 to 2 of 5xsqrt(5-x^2) ?
The integral from 1 to 2 of 5xsqrt(5-x^2) is 35/3