consider the function f(x) = e^x(sinNx) on the interval [0,1] where N is a positive integer.

a) Compute the integral from 0 to 1 of f(x). Evaluate this integral when N=5, N=10, and N=100.
B) What happens to the graph and to the value of the integral as N-->infinity? Does the graph confirm the limiting behavior of the integral's value?

look at the answer for number 13 in 7.2 in the book or the solutions manual if you have it

from what book

To compute the integral of the function f(x) = e^x(sinNx) on the interval [0,1], we can use integration by parts. However, since N is a positive integer, let's first find the antiderivative of e^x(sinNx) using substitution.

1) Antiderivative of e^x(sinNx):
Let u = Nx, then du = Ndx or dx = (1/N)du.
Now, we have:
∫ e^x(sinNx) dx = (1/N) ∫ e^(Nx)(sinNx) du

Notice that the integral ∫ e^(Nx)(sinNx) du can be seen as the imaginary part of e^(Nx) when written in terms of complex numbers. Since e^(Nx) = cos(Nx) + i*sin(Nx), the imaginary part is just sin(Nx). Thus, we get:
∫ e^x(sinNx) dx = - (1/N)Im(e^(Nx)) + C
= - (1/N)Im(cos(Nx) + i*sin(Nx)) + C
= - (1/N)sin(Nx) + C

2) Integral from 0 to 1 of f(x):
Now, we can evaluate the integral from 0 to 1 of f(x) for different values of N:
∫[0,1] f(x) dx = ∫[0,1] e^x(sinNx) dx = [- (1/N)sin(Nx)] evaluated from 0 to 1
= - (1/N)sin(N) + (1/N)sin(0)
= - (1/N)sin(N)

a) Evaluating the integral when N=5, N=10, and N=100:
For N=5: Substitute N=5 in the expression - (1/N)sin(N) to get -(1/5)sin(5).
For N=10: Substitute N=10 in the expression - (1/N)sin(N) to get -(1/10)sin(10).
For N=100: Substitute N=100 in the expression - (1/N)sin(N) to get -(1/100)sin(100).

b) Limiting behavior as N approaches infinity:
As N approaches infinity, the graph of the function oscillates more rapidly. The term sin(N) will vary between -1 and 1 for different values of N, causing the integral value to oscillate around zero. Since the integral is obtained by taking the average of the function values, as N increases, the integral will approach zero.

The graph visually confirms this limiting behavior as the oscillations become denser and closer to zero with increasing N. The area under the curve, representing the value of the integral, will tend to zero as N approaches infinity.