What is the integral from 2 to 6 of 7sqrt(x^3) ?

The integral from 2 to 6 of 7sqrt(x^3) is (7(72sqrt(6)-8sqrt(2)))/5

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integral from 2 to 6 of 7sqrt(x^3)

\int_{2}^{6} 7 x^{3} d x

\frac{7 ( 72 6 - 8 2 )}{5}

Decimal Notation

231.06937 \dots

Solution steps

\int_{2}^{6} 7 x^{3} d x

Take the constant out:

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

= 7 \cdot \int_{2}^{6} x^{3} d x

x^{3} = x^{\frac{3}{2}},

assuming

x \geq 0

= 7 \cdot \int_{2}^{6} x^{\frac{3}{2}} d x

Apply the Power Rule

: [\frac{2}{5} x^{\frac{5}{2}}]_{2}^{6}

= 7 [\frac{2}{5} x^{\frac{5}{2}}]_{2}^{6}

Compute the boundaries:

\frac{72 6 - 8 2}{5}

= 7 \cdot \frac{72 6 - 8 2}{5}

Simplify

= \frac{7 ( 72 6 - 8 2 )}{5}

f (x) = e^{x} - 3 x

\frac{\partial}{\partial y} (5 x y - 10 y)

\int \frac{1}{t + 1} d t

x \to 3 lim (\frac{x}{( 3 - x ) ^{2}})

n = 1 \sum \infty \frac{ln ( 3 n )}{n}

What is the integral from 2 to 6 of 7sqrt(x^3) ?
The integral from 2 to 6 of 7sqrt(x^3) is (7(72sqrt(6)-8sqrt(2)))/5