How do you evaluate the definite integral #int 2/(x+1)# from #[0,1]#?

Answer 1

Elijah Abram

#ln4#

This can be integrated using #int1/xdx= ln|x| + C#.

#=[2ln|x + 1|]_0^1#

#=2ln|2| - 2ln|1|#

#=2ln2=ln4#

Hopefully this helps!

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

Answer 2

Emma Aldridge

To evaluate the definite integral ∫ 2/(x+1) from 0 to 1, you can use the formula for evaluating integrals. The integral of 2/(x+1) can be found by recognizing it as the antiderivative of 2/(x+1), which is ln|x+1|. Therefore, the integral evaluates to ln|x+1| evaluated from 0 to 1. Substituting the upper limit (1) and lower limit (0) into the antiderivative function ln|x+1| and subtracting the result at the lower limit from the result at the upper limit yields the final answer.

∫ 2/(x+1) from 0 to 1 = [ln(1+1) - ln(0+1)] = [ln(2) - ln(1)] = ln(2) - 0 = ln(2).

So, the value of the definite integral is ln(2).

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.