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

The integral from-2 to-1 of 1/(x^3)+(x^3)/4 is -21/16

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

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

- \frac{21}{16}

Decimal Notation

- 1.3125

Solution steps

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

\int_{- 2}^{- 1} \frac{1}{x ^{3}} d x = - \frac{3}{8}

\int_{- 2}^{- 1} \frac{x ^{3}}{4} d x = - \frac{15}{16}

= - \frac{3}{8} - \frac{15}{16}

Simplify

- \frac{3}{8} - \frac{15}{16} : - \frac{21}{16}

= - \frac{21}{16}

\int_{3}^{4} (2 x - 5)^{4} d x

\int_{0}^{6} 3 x + 7 d x

\int_{0}^{1} (p) d p

\int_{- π}^{0} 2 x d x

\int_{- 1}^{0} 1 - 3 x d x

What is the integral from-2 to-1 of 1/(x^3)+(x^3)/4 ?
The integral from-2 to-1 of 1/(x^3)+(x^3)/4 is -21/16