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?
a) To compute the integral from 0 to 1 of f(x) = e^x(sin(Nx)) on the interval [0,1], we can use integration by parts.
Let's start by calculating the integral for general N:
∫[0,1] e^x(sin(Nx)) dx
To apply integration by parts, we'll assign u = e^x and dv = sin(Nx) dx. Taking the derivatives and integrals of these will help us simplify the integral:
du/dx = e^x
v = -1/N cos(Nx)
Using the integration by parts formula, ∫u dv = uv - ∫v du, we can rewrite the integral:
∫[0,1] e^x(sin(Nx)) dx = -e^x/N cos(Nx) - ∫[0,1] (-e^x/N cos(Nx)) dx
Now, we can apply integration by parts to the remaining integral:
u = -e^x and dv = cos(Nx) dx
du/dx = -e^x
v = 1/N sin(Nx)
Using the integration by parts formula again, we can simplify the remaining integral:
∫[0,1] (-e^x/N cos(Nx)) dx = (-e^x/N sin(Nx)) - ∫[0,1] (-e^x/N sin(Nx)) dx
Now, we have another integral to solve, but it has the same form as the original integral. We can simplify it further by applying integration by parts once again.
Using the pattern we established earlier, we'll assign u = -e^x and dv = sin(Nx) dx:
du/dx = -e^x
v = -1/N cos(Nx)
Plugging these values into the integration by parts formula, we can rewrite the remaining integral:
(-e^x/N sin(Nx)) - ∫[0,1] (-e^x/N sin(Nx)) dx = (-e^x/N sin(Nx)) - (-e^x/N cos(Nx) + ∫[0,1] (-e^x/N cos(Nx)) dx)
Simplifying further, we get:
∫[0,1] e^x(sin(Nx)) dx = -e^x/N cos(Nx) - (-e^x/N cos(Nx) + ∫[0,1] (-e^x/N cos(Nx)) dx)
Combining like terms, we have:
∫[0,1] e^x(sin(Nx)) dx = e^x/N cos(Nx) + ∫[0,1] e^x/N cos(Nx) dx
Since this integral on the right-hand side of the equation is the same as the original integral, we can substitute it with itself:
∫[0,1] e^x(sin(Nx)) dx = e^x/N cos(Nx) + ∫[0,1] e^x(sin(Nx)) dx
Now, let's solve for the value of the integral when N=5, N=10, and N=100.
When N=5:
∫[0,1] e^x(sin(5x)) dx = e^x/5 cos(5x) + ∫[0,1] e^x(sin(5x)) dx
When N=10:
∫[0,1] e^x(sin(10x)) dx = e^x/10 cos(10x) + ∫[0,1] e^x(sin(10x)) dx
When N=100:
∫[0,1] e^x(sin(100x)) dx = e^x/100 cos(100x) + ∫[0,1] e^x(sin(100x)) dx
b) As N approaches infinity, the graph of f(x) = e^x(sin(Nx)) will oscillate more rapidly. The amplitude of the oscillations will decrease because e^x remains bounded, and the period of the oscillations will decrease as N increases.
The value of the integral from 0 to 1 will also approach zero as N approaches infinity. This is because the positive and negative areas of the oscillations will cancel each other out more effectively, leading to a reduction in the net area.
Yes, the graph confirms the limiting behavior of the integral's value. As N increases, the graph becomes more tightly packed, and the area under the curve (integral) decreases.
To compute the integral from 0 to 1 of the function f(x) = e^x(sin(Nx)), we will use integration by parts along with the Fundamental Theorem of Calculus.
a) Let's start by evaluating the integral for N=5:
∫[0,1] e^x(sin(5x)) dx
Integration by parts states that ∫u dv = u v - ∫v du, where u and v are functions of x. In this case, let:
u = e^x --> du = e^x dx
dv = sin(5x) dx --> v = (-1/5) cos(5x)
Using the integration by parts formula, we have:
∫[0,1] e^x(sin(5x)) dx = (-1/5) e^x cos(5x) |[0,1] - ∫[0,1] (-1/5) e^x cos(5x) dx
Evaluating the endpoints:
(-1/5) e^1 cos(5) - (-1/5) e^0 cos(0)
Now, let's evaluate the integral for N=10 and N=100 following the same process:
For N=10:
∫[0,1] e^x(sin(10x)) dx = (-1/10) e^x cos(10x) |[0,1] - ∫[0,1] (-1/10) e^x cos(10x) dx
Evaluate the endpoints:
(-1/10) e^1 cos(10) - (-1/10) e^0 cos(0)
For N=100:
∫[0,1] e^x(sin(100x)) dx = (-1/100) e^x cos(100x) |[0,1] - ∫[0,1] (-1/100) e^x cos(100x) dx
Evaluate the endpoints:
(-1/100) e^1 cos(100) - (-1/100) e^0 cos(0)
b) As N approaches infinity, the graph of f(x) = e^x(sin(Nx)) exhibits more oscillatory behavior due to the increasing frequency of the sine function. The integral's value tends to fluctuate between positive and negative values as the area between the positive and negative lobes of the function cancels out. This behavior can be observed by comparing the values of the integrals as N increases.
In other words, as N approaches infinity, the graph becomes more "wiggly," and the integral tends to fluctuate around zero. This confirms the limiting behavior of the integral's value, as the positive and negative lobes of the function cancel each other out over the interval [0, 1].