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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 1, y^{^{'}} (1) = 0

\frac{d}{d x} (x^{4} - 4 x^{3} + 3 x^{2} - 3)

\int (x^{2} - 2 x + 5) e^{- x} d x

x \to - 2 lim (7 x^{2} + 8)

cos (- 3 x)

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