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How do you find the definite integral for: # 1/x^2# for the intervals #[0, 1]#?

Answer 1

Charles Arocha

It is

#int_0^1 1/x^2 dx=lim_(t->0^+) int_t^1 (-1/x)'dx=lim_(t->0^+) [-1/x]t^1= -1+lim(t->0^+) 1/t=oo#

Hence the integral does not converge

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Answer 2

Emma Aldridge

To find the definite integral of ( \frac{1}{x^2} ) for the intervals [0, 1], you can use the fundamental theorem of calculus.

[ \int_{0}^{1} \frac{1}{x^2} , dx ]

Using the power rule for integration, we get:

[ \int_{0}^{1} \frac{1}{x^2} , dx = \left[ -\frac{1}{x} \right]_{0}^{1} ]

[ = -\left( \frac{1}{1} \right) + \left( \frac{1}{0} \right) ]

However, at ( x = 0 ), the function ( \frac{1}{x^2} ) is undefined. Therefore, the integral does not converge.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.