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How do you find the definite integral for: #(1/x^(2))dx# for the intervals #[5, 6]#?

Answer 1

Emily Ammerman

# = 1/30#

#int_5^6 1/x^2 dx#

#= int_5^6 d/dx (-1/x) dx#

#= (-1/x)_5^6 = (1/x)_6^5#

#= 1/5 -1/6 = 1/30#

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Answer 2

Ezekiel Alexander

To find the definite integral of ( \frac{1}{x^2} ) with respect to ( x ) over the interval ([5, 6]), you evaluate the antiderivative of ( \frac{1}{x^2} ), which is ( -\frac{1}{x} ), then plug in the upper and lower bounds of the interval and take the difference.

The definite integral is:

[ \int_{5}^{6} \frac{1}{x^2} , dx = \left[-\frac{1}{x}\right]_{5}^{6} ]

[ = \left(-\frac{1}{6}\right) - \left(-\frac{1}{5}\right) ]

[ = -\frac{1}{6} + \frac{1}{5} ]

[ = \frac{5}{30} - \frac{6}{30} ]

[ = -\frac{1}{30} ]

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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.