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

The integral from 2 to x of (1+t^2) is x-2+(x^3-8)/3

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integral from 2 to x of (1+t^2)

\int_{2}^{x} (1 + t^{2}) d t

x - 2 + \frac{x ^{3} - 8}{3}

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{2}^{x} t^{2} d t = \frac{x ^{3} - 8}{3}

= x - 2 + \frac{x ^{3} - 8}{3}

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What is the integral from 2 to x of (1+t^2) ?
The integral from 2 to x of (1+t^2) is x-2+(x^3-8)/3