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

The integral from 1 to 2 of x^3+x^2-1 is 61/12

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integral from 1 to 2 of x^3+x^2-1

\int_{1}^{2} x^{3} + x^{2} - 1 d x

\frac{61}{12}

Decimal Notation

5.08333 \dots

Solution steps

\int_{1}^{2} x^{3} + x^{2} - 1 d x

Apply the Sum Rule:

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

= \int_{1}^{2} x^{3} d x + \int_{1}^{2} x^{2} d x - \int_{1}^{2} 1 d x

\int_{1}^{2} x^{3} d x = \frac{15}{4}

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

= \frac{15}{4} + \frac{7}{3} - 1

Simplify

\frac{15}{4} + \frac{7}{3} - 1 : \frac{61}{12}

= \frac{61}{12}

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What is the integral from 1 to 2 of x^3+x^2-1 ?
The integral from 1 to 2 of x^3+x^2-1 is 61/12