What is the integral from 0 to 4 of (2sqrt(x))^2 ?

The integral from 0 to 4 of (2sqrt(x))^2 is 32

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integral from 0 to 4 of (2sqrt(x))^2

\int_{0}^{4} (2 x)^{2} d x

32

Solution steps

\int_{0}^{4} (2 x)^{2} d x

Apply Integration By Parts

: [4 x^{2} - \int 4 x d x]_{0}^{4}

= [4 x^{2} - \int 4 x d x]_{0}^{4}

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

= [4 x^{2} - 2 x^{2}]_{0}^{4}

Simplify

= [2 x^{2}]_{0}^{4}

Compute the boundaries:

32

= 32

9 x, 1, 9

\frac{\partial}{\partial x} (sin^{8} (x - 4 y))

x \to 1 lim (\frac{10 x - 9 - 1}{x - 1})

\frac{d}{d x} (ln (3 x + 4))

ln (t^{2} + 6)

What is the integral from 0 to 4 of (2sqrt(x))^2 ?
The integral from 0 to 4 of (2sqrt(x))^2 is 32