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

The integral from-1 to 3 of (4x^3+7x) is 108

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integral from-1 to 3 of (4x^3+7x)

\int_{- 1}^{3} (4 x^{3} + 7 x) d x

108

Solution steps

\int_{- 1}^{3} 4 x^{3} + 7 x 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}^{3} 4 x^{3} d x + \int_{- 1}^{3} 7 x d x

\int_{- 1}^{3} 4 x^{3} d x = 80

\int_{- 1}^{3} 7 x d x = 28

= 80 + 28

Simplify

= 108

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

What is the integral from-1 to 3 of (4x^3+7x) ?
The integral from-1 to 3 of (4x^3+7x) is 108