What is the integral from-4 to 4 of (16x-(x^3)/3) ?

The integral from-4 to 4 of (16x-(x^3)/3) is 0

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integral from-4 to 4 of (16x-(x^3)/3)

\int_{- 4}^{4} (16 x - \frac{x ^{3}}{3}) d x

0

Solution steps

\int_{- 4}^{4} 16 x - \frac{x ^{3}}{3} d x

If f(x) is an odd function and continuous at:

[- a, a]

then

\int_{- a}^{a} f (x) d x = 0

Parity of

16 x - \frac{x ^{3}}{3} :

Odd

= 0

\int_{1}^{t} x e^{- x^{2}} d x

\int_{2}^{5} \frac{4}{x} d x

\int_{0}^{π} (5 + 3 sin (x)) d x

\int_{0}^{1} 4 x d x

\int_{2}^{3} (3 x^{2} + 4 x + 8) d x

What is the integral from-4 to 4 of (16x-(x^3)/3) ?
The integral from-4 to 4 of (16x-(x^3)/3) is 0