What is the integral from 0 to 7 of xsqrt(49-x^2) ?

The integral from 0 to 7 of xsqrt(49-x^2) is 343/3

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integral from 0 to 7 of xsqrt(49-x^2)

\int_{0}^{7} x 49 - x^{2} d x

\frac{343}{3}

Decimal Notation

114.33333 \dots

Solution steps

\int_{0}^{7} x 49 - x^{2} d x

Apply u-substitution

: \int_{49}^{0} - \frac{u}{2} d u

= \int_{49}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{49} - \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_{0}^{49} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{49}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{49})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{49}

Compute the boundaries:

\frac{686}{3}

= \frac{1}{2} \cdot \frac{686}{3}

Simplify

= \frac{343}{3}

\int_{0}^{2} x (x^{2} - 1)^{3} d x

\frac{\partial}{\partial x} (x^{2} y + 4 x^{4} y)

n = 3 \sum \infty \frac{1}{n ( n + 1 )}

\int \frac{a e ^{θ} + b}{a e ^{θ} - b} d θ

y = \frac{1}{1 + e ^{- x}}

What is the integral from 0 to 7 of xsqrt(49-x^2) ?
The integral from 0 to 7 of xsqrt(49-x^2) is 343/3