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

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

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integral from 1 to x of t^2

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

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

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

x y - 8 y^{2} = 4, (18, 2)

5 + 2 t cos (2 t)

f (x) = 2^{2^{x}}

\int_{1}^{6} \frac{4}{x} d x

\int 2 x + 6 d x

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