Home > Calculus > Definite and indefinite integrals

How do you evaluate the integral #int 1/x^2 dx# from 1 to #oo# if it converges?

Answer 1

Ezra Adams

#int_1^oo dx/x^2 = (-1/(2x))|_(x=1)^(x->oo)##=#

# = lim_(x->oo) (-1/(2x) )+1/2 =1/2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Arocha

To evaluate the integral ( \int_{1}^{\infty} \frac{1}{x^2} , dx ), we can use the fundamental theorem of calculus and compute the definite integral by finding the antiderivative of ( \frac{1}{x^2} ) and evaluating it at the upper and lower bounds of integration. The antiderivative of ( \frac{1}{x^2} ) is ( -\frac{1}{x} ).

So, ( \int_{1}^{\infty} \frac{1}{x^2} , dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_{1}^{b} ).

Evaluating this limit, we get: [ \lim_{b \to \infty} \left[-\frac{1}{b} + \frac{1}{1}\right] = \lim_{b \to \infty} \left[\frac{1}{1} - \frac{1}{b}\right] = 1 - 0 = 1 ]

So, the integral ( \int_{1}^{\infty} \frac{1}{x^2} , dx ) converges and its value is 1.

By signing up, you agree to our Terms of Service and Privacy Policy

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.