evaluate the following

Question

lim x to 0+ 1/x² integral from x to 3x. (e^t sin(t)) dt

Answer 1

I believe the problem looks like this:

lim(x→0+) (1/x²)∫_x^3x (e^tsint)dt

We first notice that if 0 is plugged in for x, you get an integral from 0 to 0 on the top, which is 0, and 0 on the bottom (x²) as well. This is an indeterminate form, so we must use L'Hopital's rule to evaluate the limit; take the derivative of the top (the integral) and the bottom (x²). When we derive the top, we must use the fundamental theorem of calculus:

Let's call the integrand f(t)=e^tsint, where F(t) would be the antiderivative of this expression. Then, plugging in the limits, the integral would equal F(3x)-F(x). Then, taking the derivative of this, we must use the chain rule (pulling out a 3), giving 3f(3x)-f(x) (where f is the original function).

Then derive the bottom (x²), giving 2x. The limit is now reduced to:

lim(x→0+) (3f(3x)-f(x))/(2x)

If we plug in zero again, we still get an in indeterminate form, 0/0. Thus, we must perform L'Hopital's rule again. Taking the derivative of the top, using the chain rule again, we get 3*3f '(3x)-f '(x)=9f '(x)-f '(x) (where f ' is the derivative of the original function). Taking the derivative of the bottom (2x), we get 2. The limit now becomes:

lim(x→0+) (9f '(3x)-f '(x))/2

Now, the limit will have a determinate form (because the bottom is nonzero). We simply need to evaluate the expression, so first we need to find f ', then plug in the appropriate values:

Taking the derivative of f(t)=e^tsint, using the product rule, we get f '(t)=e^tsint+e^tcost

So the top expression becomes 9(e^3xsin3x+e^3xcos3x)-(e^xsinx+e^xcosx)

Now, plugging in 0, we get (9(e⁰sin0+e⁰cos0)-(e⁰sin0+e⁰cos0))/2=(9(0+1)-(0+1))/2=8/2=4