What is the integral from 1 to x^3 of t^2 ?

The integral from 1 to x^3 of t^2 is (x^9-1)/3

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integral from 1 to x^3 of t^2

\int_{1}^{x^{3}} t^{2} d t

\frac{x ^{9} - 1}{3}

Solution steps

\int_{1}^{x^{3}} t^{2} d t

Apply the Power Rule

: [\frac{t ^{3}}{3}]_{1}^{x^{3}}

= [\frac{t ^{3}}{3}]_{1}^{x^{3}}

Compute the boundaries:

\frac{x ^{9}}{3} - \frac{1}{3}

= \frac{x ^{9}}{3} - \frac{1}{3}

Simplify

= \frac{x ^{9} - 1}{3}

x \to 0 lim (\frac{e ^{- 5 x} - 1}{x})

\frac{d}{d x} (cos (5 x^{2} - x))

f (x) = 2 x^{2} + 3 x, at x = - 2

f (x) = e^{3 e^{x}}

\int (1 + sec (x))^{2} d x

What is the integral from 1 to x^3 of t^2 ?
The integral from 1 to x^3 of t^2 is (x^9-1)/3