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

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

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

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

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

Solution steps

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

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

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

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

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