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

The integral from 0 to 2 of t^3(2+t^4)^3 is 6560

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

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

6560

Solution steps

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

Apply u-substitution

= \int_{2}^{18} \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}^{18} u^{3} d u

Apply the Power Rule

= \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{18}

Compute the boundaries:

26240

= \frac{1}{4} \cdot 26240

Simplify

= 6560

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t an g e n t (3 x + 72 at (3.9))

x \to 1 lim (3 - x)

\frac{d y}{d x} (y) = (y)^{2} \cdot e^{3 x}

What is the integral from 0 to 2 of t^3(2+t^4)^3 ?
The integral from 0 to 2 of t^3(2+t^4)^3 is 6560