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Evaluate the definite integrals?
Hey there I'd like a step by step process if possible.
a)
e^4
f thing dx/ xsqrt(lnx)
e
b)
4
f thing xdx/ sqrt(1+2x)
0
I thought of maybe doing the substitution method as that's what we're covering in class but that only seems to apply to indefinite integrals. So I'm kind of stuck.
Thanks.
2 Answers
- Anonymous8 years agoFavorite Answer
∫dx / (x √(lnx) )
let u = lnx
du = dx/x
∫1/√(u) du
2√(u)
2√(lnx)
2√(lne^4) - 2√(lne) = 2
∫x dx/√(1 + 2x)
let u = 1 + 2x
x = (u-1)/2
dx = (1/2) du
(1/2) ∫[(u-1)/2 /√(u)] du
(1/4) ∫u^(1/2) - u^(-1/2) du
(1/4) [(2/3) u^(3/2) - 2u^(1/2)]
(1/4) [(2/3) (1+2x)^(3/2) - 2(1+2x)^(1/2)]
(1/4) [(2/3) (1+2(4))^(3/2) - 2(1+2(4))^(1/2)] - (1/4) [(2/3) (1+2(0))^(3/2) - 2(1+2(0))^(1/2)]
= 10/3
- Anonymous8 years ago
a. Do a u-substituation where u = lnx, and du = 1/x. You have to change your bounds as well to accompany the u-substiution, so plug in your bound in the u = lnx. This will change your bound to 1-4, and change your equation to du/ sqrt(u) = du/ (u)^1/2. You then need to integrate, which should give you 2*sqrt(u) over the interval 1-4. Now just plug in bound to evaluate, so 2*sqrt(4) - 2*sqrt(1) = 4-2 = 2.
b. Sorry, this one is kind of weird so i'm not quite sure i know how to do it, so i think someone else should answer.
Source(s): Calculus 2