Find the area between the graph of f(x)= x(e^-x^2) and the x-axis for the interval (0, ∞). Your work must include the proper notation and show the antiderivative. If the integral diverges, show why.
∫xe^(-x^2) dx = -1/2 e^(-x^2)
so, integrating from 0 to ∞, that is
-1/2 [ lim(x->∞) e^(-x^2) - 1] = -1/2 [0 - 1] = 1/2
To find the area between the graph of a function and the x-axis, we need to evaluate the definite integral of the absolute value of the function over the given interval. In this case, we want to find the area between the graph of f(x) = x(e^(-x^2)) and the x-axis for the interval (0, ∞).
First, let's find the antiderivative of f(x):
∫x(e^(-x^2)) dx
We can rewrite this integral using substitution. Let u = -x^2, then du/dx = -2x, and dx = du / -2x.
Now, substituting, we have:
∫x(e^u) (du / -2x)
The x in the numerator cancels out with the x in the denominator, and we are left with:
-1/2 ∫e^u du
Integrating e^u with respect to u gives us e^u:
-1/2 e^u
Now, let's evaluate the antiderivative for our definite integral over the interval (0, ∞):
∫(0 to ∞) x(e^-x^2) dx = lim as b→∞ [-1/2 e^(-x^2)] evaluated from 0 to b
Using the limit as b approaches infinity, we have:
lim as b→∞ [-1/2 e^(-b^2) - (-1/2 e^(-0^2))]
Simplifying further:
lim as b→∞ -1/2 (e^(-b^2) - 1)
Now, let's analyze the behavior of e^(-b^2) as b approaches infinity. Exponential functions with negative exponents decay rapidly as the value inside the exponent (in this case, -b^2) becomes increasingly large. Therefore, as b approaches infinity, e^(-b^2) approaches zero.
Therefore, the integral does not converge, and the area between the graph of f(x) = x(e^(-x^2)) and the x-axis for the interval (0, ∞) is undefined.
To find the area between the graph of the function f(x) = x(e^-x^2) and the x-axis for the interval (0, ∞), we can use calculus and integration.
First, let's find the antiderivative of the function f(x) = x(e^-x^2). We can do this by using integration by substitution. Let's assume u = -x^2 and find the derivative of u with respect to x.
du/dx = -2x
Now, we can rewrite the function f(x) as:
f(x) = x(e^-x^2)
Since du/dx = -2x, we can rewrite the function in terms of u:
f(x) = -1/2 * (e^u)
Next, let's integrate the function in terms of u:
∫ -1/2 * (e^u) du
Using the power rule of integration, the antiderivative of -1/2 * (e^u) is:
-1/2 * (e^u) / (-1/2) = 2e^u
Replacing u with -x^2, we get:
2e^(-x^2)
To find the area between the graph and the x-axis, we want to evaluate the definite integral of the absolute value of the function between the interval (0, ∞):
∫[0, ∞] |f(x)| dx = ∫[0, ∞] |x(e^-x^2)| dx = ∫[0, ∞] x(e^-x^2) dx
To evaluate the definite integral, we can use integration by parts. Let u = x and dv = e^-x^2 dx. Then, du = dx and v can be found by integrating dv:
v = ∫ e^-x^2 dx
The integral on the right side does not have a closed-form solution, meaning it cannot be expressed in terms of elementary functions. Therefore, the integral ∫[0, ∞] x(e^-x^2) dx diverges.
Hence, the area between the graph of f(x) = x(e^-x^2) and the x-axis for the interval (0, ∞) is undefined or infinite.