What is the integral from 0 to 1 of t^3(5+t^4)^3 ?

The integral from 0 to 1 of t^3(5+t^4)^3 is 671/16

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integral from 0 to 1 of t^3(5+t^4)^3

\int_{0}^{1} t^{3} (5 + t^{4})^{3} d t

\frac{671}{16}

Decimal

41.9375

Solution steps

\int_{0}^{1} t^{3} (5 + t^{4})^{3} d t

Expand

t^{3} (5 + t^{4})^{3} : 125 t^{3} + 75 t^{7} + 15 t^{11} + t^{15}

= \int_{0}^{1} 125 t^{3} + 75 t^{7} + 15 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} 125 t^{3} d t + \int_{0}^{1} 75 t^{7} d t + \int_{0}^{1} 15 t^{11} d t + \int_{0}^{1} t^{15} d t

\int_{0}^{1} 125 t^{3} d t = \frac{125}{4}

\int_{0}^{1} 75 t^{7} d t = \frac{75}{8}

\int_{0}^{1} 15 t^{11} d t = \frac{5}{4}

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

= \frac{125}{4} + \frac{75}{8} + \frac{5}{4} + \frac{1}{16}

Simplify

\frac{125}{4} + \frac{75}{8} + \frac{5}{4} + \frac{1}{16} : \frac{671}{16}

= \frac{671}{16}

\int_{- a}^{a} x d x

\int_{- 3}^{0} x^{2} + 3 x d x

\int_{0}^{\frac{3}{8}} \frac{1}{64 x ^{2} + 9} d x

\int_{- 1}^{1} (- x^{2} ∣ x ∣) d x

\int_{0}^{1} x^{2} \cdot e^{x} d x

What is the integral from 0 to 1 of t^3(5+t^4)^3 ?
The integral from 0 to 1 of t^3(5+t^4)^3 is 671/16