What is the integral from 0 to 1 of t^3(2+t^4)^3 ?

The integral from 0 to 1 of t^3(2+t^4)^3 is 65/16

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integral from 0 to 1 of t^3(2+t^4)^3

\int_{0}^{1} t^{3} (2 + t^{4})^{3} d t

\frac{65}{16}

Decimal

4.0625

Solution steps

\int_{0}^{1} t^{3} (2 + t^{4})^{3} d t

Apply u-substitution

: \int_{2}^{3} \frac{u ^{3}}{4} d u

= \int_{2}^{3} \frac{u ^{3}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{2}^{3} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2}^{3}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{3}

Compute the boundaries:

\frac{65}{4}

= \frac{1}{4} \cdot \frac{65}{4}

Simplify

= \frac{65}{16}

\int_{\infty}^{1} \frac{1}{x + 3} d x

\int_{- 1}^{1} ∣ 2 - 4 x ∣ d x

\int_{0}^{2} (4 - y^{2}) d y

\int_{0}^{11} ∣ 9 - x ∣ d x

\int_{0}^{1} y^{2} (1 - y^{\frac{3}{2}}) d y

What is the integral from 0 to 1 of t^3(2+t^4)^3 ?
The integral from 0 to 1 of t^3(2+t^4)^3 is 65/16