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

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

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

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

0

Solution steps

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

Odd

= 0

\int \frac{x ^{2} - x + 2}{3 x ^{3} - 2 x ^{2} + 6 x - 4} d x

y = \frac{x ^{2}}{2} + 2, [- 3, 0]

\frac{d}{d x} ((3 x + 6)^{2})

\frac{d}{d x} (- \frac{11}{x ^{12}})

y = - x^{2} - 2 x + 3, y = - x + 1

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