Calculus/Trouble with integration
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
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)