Home > Calculus > Definite and indefinite integrals

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

Answer 1

Elijah Abram

#4(sqrt2-1)#.

Let #1+x=t^2 rArr dx=2tdt#

Also, #x=0 rArr t^2=1 rArr t=1, &, x=1 rArr t=sqrt2#. Hence,

#I=int_0^1 2/sqrt(1+x)dx=int_1^sqrt2 2/sqrt(t^2)*2tdt=4int_1^sqrt2 dt#

#=4[t]_1^sqrt2#

#=4(sqrt2-1)#.

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

Answer 2

Scarlett Allison

To evaluate the definite integral ∫(2/√(1+x)) dx from 0 to 1, you can use the substitution method. Let u = 1 + x. Then, du = dx.

So, when x = 0, u = 1, and when x = 1, u = 2.

The integral becomes: ∫(2/√u) du from 1 to 2.

This integrates to: 4√u evaluated from 1 to 2.

Substituting the limits, we get: 4√2 - 4√1.

Which simplifies to: 4√2 - 4.

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.