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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 0

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 1}^{0} - \frac{4}{x ^{3}} d x + \int_{0}^{2} - \frac{4}{x ^{3}} d x

\int_{- 1}^{0} - \frac{4}{x ^{3}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{t} e^{4 t} d t

\int_{1}^{2} x^{2} - 5 x + 6 d x

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

\int_{0}^{4} (4 - t) (t) d t

\int_{0}^{3} (2 x^{2} + 3) d x

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