What is the integral from 4 to 0 of x^{1/2} ?

The integral from 4 to 0 of x^{1/2} is -16/3

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integral from 4 to 0 of x^{1/2}

\int_{4}^{0} x^{\frac{1}{2}} d x

- \frac{16}{3}

Solution steps

\int_{4}^{0} x^{\frac{1}{2}} d x

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

= - \int_{0}^{4} x^{\frac{1}{2}} d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{4}

= - [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= - \frac{16}{3}

\int_{0}^{1} (4 x - 3) d x

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

\int_{1}^{3} \frac{2}{8 + 4 x} d x

\int_{2}^{\infty} \frac{1}{x ^{3} - 8} d x

\int_{- 9}^{\infty} - 4 e^{- 4 x} d x

What is the integral from 4 to 0 of x^{1/2} ?
The integral from 4 to 0 of x^{1/2} is -16/3