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

The integral from 0 to 1 of t^3(1+t^4)^3 is 15/16

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

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

\frac{15}{16}

Decimal Notation

0.9375

Solution steps

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

Expand

t^{3} (1 + t^{4})^{3} : t^{3} + 3 t^{7} + 3 t^{11} + t^{15}

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

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

\int_{0}^{1} 3 t^{7} d t = \frac{3}{8}

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

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

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

Simplify

\frac{1}{4} + \frac{3}{8} + \frac{1}{4} + \frac{1}{16} : \frac{15}{16}

= \frac{15}{16}

\frac{d y}{d x} = - 10 x^{4} + 8 x^{3}

f (x) = 7 x - x, (4, 26)

f (x) = (\frac{x + 11}{x - 11})^{6}

\int x^{3} e^{4 x^{2}} d x

\frac{d}{d x} (\frac{2 sin ( x )}{sec ( x )})

What is the integral from 0 to 1 of t^3(1+t^4)^3 ?
The integral from 0 to 1 of t^3(1+t^4)^3 is 15/16