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

The integral from 0 to 2 of t^3(3+t^4)^3 is 8140

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

\int_{0}^{2} t^{3} (3 + t^{4})^{3} d t

8140

Solution steps

\int_{0}^{2} t^{3} (3 + t^{4})^{3} d t

Expand

t^{3} (3 + t^{4})^{3} : 27 t^{3} + 27 t^{7} + 9 t^{11} + t^{15}

= \int_{0}^{2} 27 t^{3} + 27 t^{7} + 9 t^{11} + t^{15} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2} 27 t^{3} d t + \int_{0}^{2} 27 t^{7} d t + \int_{0}^{2} 9 t^{11} d t + \int_{0}^{2} t^{15} d t

\int_{0}^{2} 27 t^{3} d t = 108

\int_{0}^{2} 27 t^{7} d t = 864

\int_{0}^{2} 9 t^{11} d t = 3072

\int_{0}^{2} t^{15} d t = 4096

= 108 + 864 + 3072 + 4096

Simplify

= 8140

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

f (x) = 2 - x

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

\int_{300}^{360} \frac{2400}{x ^{2}} d x

x \frac{d y}{d x} + y = y^{- 2}

What is the integral from 0 to 2 of t^3(3+t^4)^3 ?
The integral from 0 to 2 of t^3(3+t^4)^3 is 8140