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

The integral from 0 to 1 of t^3(3+t^4)^3 is 175/16

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

\int_{0}^{1} t^{3} (3 + t^{4})^{3} d t

\frac{175}{16}

Decimal Notation

10.9375

Solution steps

\int_{0}^{1} 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}^{1} 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}^{1} 27 t^{3} d t + \int_{0}^{1} 27 t^{7} d t + \int_{0}^{1} 9 t^{11} d t + \int_{0}^{1} t^{15} d t

\int_{0}^{1} 27 t^{3} d t = \frac{27}{4}

\int_{0}^{1} 27 t^{7} d t = \frac{27}{8}

\int_{0}^{1} 9 t^{11} d t = \frac{3}{4}

\int_{0}^{1} t^{15} d t = \frac{1}{16}

= \frac{27}{4} + \frac{27}{8} + \frac{3}{4} + \frac{1}{16}

Simplify

\frac{27}{4} + \frac{27}{8} + \frac{3}{4} + \frac{1}{16} : \frac{175}{16}

= \frac{175}{16}

f (x) = 12 - x

t \to 0 lim (\frac{sin ( 3 t )}{t})

y = 3 x + 6 cos (x), (\frac{π}{3}, π + 3)

\frac{x ^{2} - 1}{x ^{2} + x + 1}

\int e^{x} + 1 d x

What is the integral from 0 to 1 of t^3(3+t^4)^3 ?
The integral from 0 to 1 of t^3(3+t^4)^3 is 175/16