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

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

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

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

0

Solution steps

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

Odd

= 0

\int_{0}^{1} 1 + 4 t^{2} d t

\int_{1}^{2} \frac{1 + ln ( x )}{x} d x

\int_{4}^{3} sin (t) d t

\int_{0}^{1} 3 (1 - y)^{2} d y

\int_{0}^{1} e^{- s x} d x

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