What is the integral from-1 to 1 of t^3(2+t^4)^3 ?

The integral from-1 to 1 of t^3(2+t^4)^3 is 0

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integral from-1 to 1 of t^3(2+t^4)^3

\int_{- 1}^{1} t^{3} (2 + t^{4})^{3} d t

0

Solution steps

\int_{- 1}^{1} t^{3} (2 + t^{4})^{3} d t

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

[- a, a]

then

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

Parity of

t^{3} (2 + t^{4})^{3} :

Odd

= 0

\int_{0}^{1} sinh (4 x) (x^{2} + 5) d x

\int_{0}^{1} \frac{1}{x ^{2} - 2 x - 3} d x

\int_{- 1}^{4} x^{2} d x

\int_{0}^{3} (2 - x) d x

\int_{- 2}^{2} x^{3} + x d x

What is the integral from-1 to 1 of t^3(2+t^4)^3 ?
The integral from-1 to 1 of t^3(2+t^4)^3 is 0