What is the integral from a to b of x^2 1/(b-a) ?

The integral from a to b of x^2 1/(b-a) is (b^2+ba+a^2)/3

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integral from a to b of x^2 1/(b-a)

\int_{a}^{b} x^{2} \frac{1}{b - a} d x

\frac{b ^{2} + ba + a ^{2}}{3}

Solution steps

\int_{a}^{b} x^{2} \frac{1}{b - a} d x

Take the constant out:

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

= \frac{1}{b - a} \cdot \int_{a}^{b} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{a}^{b}

= \frac{1}{b - a} [\frac{x ^{3}}{3}]_{a}^{b}

Simplify

\frac{1}{b - a} [\frac{x ^{3}}{3}]_{a}^{b} : \frac{[ \frac{x ^{3}}{3} ] _{a}^{b}}{b - a}

= \frac{[ \frac{x ^{3}}{3} ] _{a}^{b}}{b - a}

Compute the boundaries:

\frac{b ^{3}}{3} - \frac{a ^{3}}{3}

= \frac{\frac{b ^{3}}{3} - \frac{a ^{3}}{3}}{b - a}

Simplify

= \frac{b ^{2} + ba + a ^{2}}{3}

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What is the integral from a to b of x^2 1/(b-a) ?
The integral from a to b of x^2 1/(b-a) is (b^2+ba+a^2)/3