What is the integral from 0 to 5r of 2(25r^2-y^2) ?

The integral from 0 to 5r of 2(25r^2-y^2) is 2(-(125r^3)/3+125r^3)

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integral from 0 to 5r of 2(25r^2-y^2)

\int_{0}^{5 r} 2 (25 r^{2} - y^{2}) d y

2 (- \frac{125 r ^{3}}{3} + 125 r^{3})

Solution steps

\int_{0}^{5 r} 2 (25 r^{2} - y^{2}) d y

Take the constant out:

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

= 2 \cdot \int_{0}^{5 r} - y^{2} + 25 r^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int_{0}^{5 r} y^{2} d y + \int_{0}^{5 r} 25 r^{2} d y)

\int_{0}^{5 r} y^{2} d y = \frac{125 r ^{3}}{3}

\int_{0}^{5 r} 25 r^{2} d y = 125 r^{3}

= 2 (- \frac{125 r ^{3}}{3} + 125 r^{3})

\int \frac{1}{y ^{3} + 1} d y

y = 6 tan (x), at x = \frac{π}{3}

\int \frac{10 x ^{2} - 9 x + 20}{x ^{3} + 4 x} d x

x y^{^{'}} + x y - y^{^{'}} - 2 y = 0

\int - 2 tan (2 x) d x

What is the integral from 0 to 5r of 2(25r^2-y^2) ?
The integral from 0 to 5r of 2(25r^2-y^2) is 2(-(125r^3)/3+125r^3)