What is the integral from 0 to 2 of pi(sqrt(8x))^2 ?

The integral from 0 to 2 of pi(sqrt(8x))^2 is pi16

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integral from 0 to 2 of pi(sqrt(8x))^2

\int_{0}^{2} π (8 x)^{2} d x

π 16

Decimal

50.26548 \dots

Solution steps

\int_{0}^{2} π (8 x)^{2} d x

Take the constant out:

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

= π \cdot \int_{0}^{2} (8 x)^{2} d x

Apply Integration By Parts

: [8 x^{2} - \int 8 x d x]_{0}^{2}

= π [8 x^{2} - \int 8 x d x]_{0}^{2}

\int 8 x d x = 4 x^{2}

= π [8 x^{2} - 4 x^{2}]_{0}^{2}

Simplify

= π [4 x^{2}]_{0}^{2}

Compute the boundaries:

16

= π 16

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What is the integral from 0 to 2 of pi(sqrt(8x))^2 ?
The integral from 0 to 2 of pi(sqrt(8x))^2 is pi16