so I didn't find an answer online, so I would like to ask which answer is correct:
$$\int_{-4}^4 x^3 \,dx = 0 $$
or
$$\int_{-4}^4 x^3 \,dx = 64 $$
Thank you!
Asked
Active
Viewed 46 times
-3
-
Hint: $x^3$ is odd. – poyea Jun 14 '18 at 11:05
-
Yea, that's why I'm asking it, however I doubt the area from -4 to 0 "cancels" the area from 0 to 4, but not sure, that's why I'm asking – Qemikal Jun 14 '18 at 11:07
3 Answers
1
Hint:
$f(x)=x^3$ is an odd function. This means that $f(-x)=-f(x)$ and we have:
$$ \int_{-a}^a f(x)dx=\int_{-a}^0 f(x)dx+\int_{0}^a f(x)dx=\int_{0}^a f(-x)dx+\int_{0}^a f(x)dx=-\int_{0}^a f(x)dx+\int_{0}^a f(x)dx $$
Emilio Novati
- 62,675
0
$f(-x)=-f(x)$
$$ \int_{0}^{4}f\left(x\right)\text{d}x\underset{x \rightarrow -x}{=}\int_{0}^{-4}f\left(-x\right)\left(-\text{d}x\right)=-\int_{-4}^{0}f\left(x\right)\text{d}x $$ So it's null summung over $-4$ to $4$.
Atmos
- 7,369
