What is the integral from 3 to 4 of xsqrt(25-x^2) ?

The integral from 3 to 4 of xsqrt(25-x^2) is 37/3

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integral from 3 to 4 of xsqrt(25-x^2)

\int_{3}^{4} x 25 - x^{2} d x

\frac{37}{3}

Decimal Notation

12.33333 \dots

Solution steps

\int_{3}^{4} x 25 - x^{2} d x

Apply u-substitution

: \int_{16}^{9} - \frac{u}{2} d u

= \int_{16}^{9} - \frac{u}{2} d u

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

= - \int_{9}^{16} - \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_{9}^{16} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

Compute the boundaries:

\frac{74}{3}

= \frac{1}{2} \cdot \frac{74}{3}

Simplify

= \frac{37}{3}

\int_{0}^{\frac{π}{2}} \frac{1}{2 cos ( x ) + 3} d x

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

\int_{0}^{\frac{π}{2}} tan^{2} (θ) d θ

\int_{1}^{4} 5 d x

\int_{1}^{4} 2 d x

What is the integral from 3 to 4 of xsqrt(25-x^2) ?
The integral from 3 to 4 of xsqrt(25-x^2) is 37/3