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

n = 0 \sum \infty \frac{4}{7 ^{n - 1}}

\frac{\partial}{\partial x} (- sin (x y))

y = 0.5 x - 3

\int_{0}^{l n (4)} e^{x} - e^{- 2 x} d x

\int \frac{3 x ^{2}}{( 2 x ^{3} + 1 ) ^{4}} 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