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

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

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

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

0

Solution steps

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

Odd

= 0

\int (0.3 y^{4} - 8 y^{- 3} + 2) d y

x \to - 1 + lim (x)

\frac{d}{d x} (\frac{( 1 + x ^{2} + x ^{4} )}{1 + x + x ^{2}})

y^{^{'}} - \frac{9}{x} y = \frac{y ^{5}}{x ^{20}}

y^{^{''}} + 16 y = sin (4 t)

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