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How do you evaluate the definite integral #int (1+y^2)/y dy# from #[1,2]#?

Answer 1

Scarlett Allison

#ln2+3/2#

#int_1^2(1+y^2)/ydy#

#=int_1^2(1/y+y^2/y)dy=int_1^2(1/y+y)dy#

#=[lny+y^2/2]_1^2#

#=[lny+y^2/2]^2-[lny+y^2/2]_1#

#=[ln2+2^2/2]-[cancel(ln1)^0+1^2/2]#

#=ln2+2-1/2#

#=ln2+3/2#

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Answer 2

Samantha Adkison

To evaluate the definite integral ∫(1+y^2)/y dy from 1 to 2, you can use the properties of integrals and apply the fundamental theorem of calculus. First, rewrite the integrand as (1/y) + y. Then integrate each term separately. The integral of 1/y is ln|y|, and the integral of y is (1/2)y^2. Evaluate each integral at the upper and lower limits of integration (2 and 1, respectively), and subtract the result of the lower limit from the upper limit.

So, ∫(1+y^2)/y dy from 1 to 2 = [ln|2| + (1/2)(2)^2] - [ln|1| + (1/2)(1)^2] = [ln(2) + 2] - [0 + 1/2] = ln(2) + 3/2.

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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.