What is the integral from 3 to x^3 of t^4 ?

The integral from 3 to x^3 of t^4 is (x^{15}-243)/5

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integral from 3 to x^3 of t^4

\int_{3}^{x^{3}} t^{4} d t

\frac{x ^{15} - 243}{5}

Solution steps

\int_{3}^{x^{3}} t^{4} d t

Apply the Power Rule

: [\frac{t ^{5}}{5}]_{3}^{x^{3}}

= [\frac{t ^{5}}{5}]_{3}^{x^{3}}

Compute the boundaries:

\frac{x ^{15}}{5} - \frac{243}{5}

= \frac{x ^{15}}{5} - \frac{243}{5}

Simplify

= \frac{x ^{15} - 243}{5}

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

\frac{\partial}{\partial y} (\frac{( 2 x + y )}{z})

f (x) = \frac{4}{x}, (4, 2)

4 e^{4 x}

6 x + 5

What is the integral from 3 to x^3 of t^4 ?
The integral from 3 to x^3 of t^4 is (x^{15}-243)/5