What is the integral from 0 to 1 of 6sqrt(x) ?

The integral from 0 to 1 of 6sqrt(x) is 4

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integral from 0 to 1 of 6sqrt(x)

\int_{0}^{1} 6 x d x

4

Solution steps

\int_{0}^{1} 6 x d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} x d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{1}

= 6 [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= 6 \cdot \frac{2}{3}

Simplify

= 4

\int_{0}^{t^{2}} u d u

\int_{2}^{3} \frac{( 2 x - 3 )}{4 x - x ^{2}} d x

\int_{\frac{1}{12}}^{\frac{1}{4}} \frac{2}{t + 4 t t} d t

\int_{4}^{5} x^{2} + 2 x - 11 d x

\int_{0}^{4} (x^{2} - 5) d x

What is the integral from 0 to 1 of 6sqrt(x) ?
The integral from 0 to 1 of 6sqrt(x) is 4