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Hi Amos!

These two problems are giving me very much trouble.

1. find the value of "a" for the integral

(x e^x) dx =3a , lower limit: 1, upper limit: ln (a)

2. Find the value of the integral

e^sqrt(ax) d(sqrt x) , lower limit: 0, upper limit: 4/a

Thank you!

The second problem is easier, but the first one involves more algebra that respects logarithm.

1) Assuming you know how to integrate by parts, the first one ha s function whose anti-derivative (or integral) is F(x)= xe^x - e^x.

So the definite integral by the fundamental theorem of calculus becomes F(ln a) - F(1). i.e. we have the equation:

F(ln a) - F(1) = 3a

=> (ln a -1) a - (1e^1 - e^1) = 3a

=> (ln a -1) a = 3a

=> (ln a -4) a = 0 (I hope you understand the algebra needed to get here)

so either a = 0 or ln a = 4

i.e a = 0 or a = e^4.

But we want a limit from 1 to ln(a), so a cannot be zero, that leaves us with the asnwer a = e^4.

2) If you do not understand what d(sqrt x) means, please refer to integration by substitution.

Doing so, we will get the integral of e^sqrt(ax) d(sqrt x) = G(x) = e^sqrt(ax) / sqrt(a).

So again using the fundamental theorem of calculus, the definite integral becomes G(4/a) - G(0)

G(4/a) = e^2/sqrt(a) and G(0) = 1.

Hence the integral e^sqrt(ax) d(sqrt x) , lower limit: 0, upper limit: 4/a is equal to

(e^2 - 1) / sqrt(a)

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Calculus

Answers by Expert:

I can answer all calculus question. I am a Math Lecturer and I teach Math in a College, usually Calculus 1, Calculus 2 and Calculus 3 and Linear Algebra.

I have been teaching in this college for more than 12 years. Prior to that, I taught for three years as Visiting Assistant Professor.**Education/Credentials**

I got my doctorate from Univ. of Rochester in Algebraic Toppology. I got a MA from Univ of Rochester, an MA from York Univ. in Toronto and almost did my M.Sc in the National Univ. of Singapore.