What is the integral from-1 to 3 of (x^3)/4+x^2-5/4 ?

The integral from-1 to 3 of (x^3)/4+x^2-5/4 is 28/3

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integral from-1 to 3 of (x^3)/4+x^2-5/4

\int_{- 1}^{3} \frac{x ^{3}}{4} + x^{2} - \frac{5}{4} d x

\frac{28}{3}

Decimal Notation

9.33333 \dots

Solution steps

\int_{- 1}^{3} \frac{x ^{3}}{4} + x^{2} - \frac{5}{4} 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}^{3} \frac{x ^{3}}{4} d x + \int_{- 1}^{3} x^{2} d x - \int_{- 1}^{3} \frac{5}{4} d x

\int_{- 1}^{3} \frac{x ^{3}}{4} d x = 5

\int_{- 1}^{3} x^{2} d x = \frac{28}{3}

\int_{- 1}^{3} \frac{5}{4} d x = 5

= 5 + \frac{28}{3} - 5

Simplify

= \frac{28}{3}

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What is the integral from-1 to 3 of (x^3)/4+x^2-5/4 ?
The integral from-1 to 3 of (x^3)/4+x^2-5/4 is 28/3