What is the integral from-1 to b of x^2(7-x) ?

The integral from-1 to b of x^2(7-x) is (7(b^3+1))/3-(b^4-1)/4

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integral from-1 to b of x^2(7-x)

\int_{- 1}^{b} x^{2} (7 - x) d x

\frac{7 ( b ^{3} + 1 )}{3} - \frac{b ^{4} - 1}{4}

Solution steps

\int_{- 1}^{b} x^{2} (7 - x) d x

Expand

x^{2} (7 - x) : 7 x^{2} - x^{3}

= \int_{- 1}^{b} 7 x^{2} - x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{b} 7 x^{2} d x - \int_{- 1}^{b} x^{3} d x

\int_{- 1}^{b} 7 x^{2} d x = \frac{7 ( b ^{3} + 1 )}{3}

\int_{- 1}^{b} x^{3} d x = \frac{b ^{4} - 1}{4}

= \frac{7 ( b ^{3} + 1 )}{3} - \frac{b ^{4} - 1}{4}

What is the integral from-1 to b of x^2(7-x) ?
The integral from-1 to b of x^2(7-x) is (7(b^3+1))/3-(b^4-1)/4