What is the integral from 0 to x of (t^3+t+1) ?

The integral from 0 to x of (t^3+t+1) is (x^4)/4+(x^2)/2+x

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integral from 0 to x of (t^3+t+1)

\int_{0}^{x} (t^{3} + t + 1) d t

\frac{x ^{4}}{4} + \frac{x ^{2}}{2} + x

Solution steps

\int_{0}^{x} t^{3} + t + 1 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_{0}^{x} t^{3} d t + \int_{0}^{x} t d t + \int_{0}^{x} 1 d t

\int_{0}^{x} t^{3} d t = \frac{x ^{4}}{4}

\int_{0}^{x} t d t = \frac{x ^{2}}{2}

\int_{0}^{x} 1 d t = x

= \frac{x ^{4}}{4} + \frac{x ^{2}}{2} + x

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What is the integral from 0 to x of (t^3+t+1) ?
The integral from 0 to x of (t^3+t+1) is (x^4)/4+(x^2)/2+x