82. A family of exponentials The curves y = x * e^(-a * x) are...

Question

82. A family of exponentials The curves y = x * e^(-a * x) are shown in the figure for a = 1, 2, and 3.

c. Find the area of the region bounded by y = x * e^(-a * x) and the x-axis on the interval [0, b]. Because this area depends on a and b, we call it A(a, b).

Graph showing three curves of y = x * e^(-a * x) for a = 1, 2, and 3, all approaching the x-axis.

Accepted Answer

Welcome back, everyone. Find the area enclosed by the shaded region in the given figure. In this problem, we want to identify the area enclosed by the x-axis from the origin of X equals 0 up to the point A, and the curve that is above the X axis is Y equals X multiplied by e to the power of negative X divided by 2. Because the shaded region is above the x-axis, we can integrate directly and show that A. The area is equal to the integral from 0 to A of the equation of the curve, so X multiplied by e to the power of negative X divided by 2D X. So the whole idea in this problem is to integrate this function. We're going to use integration by parts, and let's begin by identifying an indefinite integral in the form of integral of X to the power of negative X divided by 2D X. We're going to set U equals X, which means that DU is equal to DX. And DV is going to be the remaining part E to the power of negative X divided by 2D X meaning V is the integral of the V and integrating this expression, we get negative2 E to the power of negative X divided by 2, right? So now we're using the integration by parts formula, integral of UDV equals UV minus the integral of VDU, we can show that the integral of X eats the power of negative X divided by 2D X is equal to UV. So we get -2 X E to the power of negative X divided by 2 minus the integral of VDU, right? So, essentially what we have to do is Take the integral of. We can write down V, which is -2. E to the power of negative X divided by 2. And the U is DX. So now let's continue our equation. We get -2 X. E to the power of negative X divided by 2. We can factor out the constant, so we get + 2 minus -2 gives us + 2. Integral of E to the power of negative X divided by 2D X and we already know that this integral is equal to V. So we get -2 XE to the power of negative X divided by 2 + 2 multiplied by -2 E to the power of negative X divided by 2 plus a constant of integration C because this is an indefinite integral, right? So now with simplifying, we get -2 XE to the power of negative X divided by 2 minus 4E to the power of negative X divided by 2 + C. and we can also factor to show that this is 2 to the power of negative X divided by 2. Andres we're going to have X. Plus 2 plus a constant of integration C. So now considering the actual area we know that it is equal to this integral evaluated from 0 to A. In other words, we want to evaluate -2 to the power of negative X divided by 2. Multiplied by X + 2 from 0 to A. Substituting a we get -2 E to the power of negative a divided by 2 multiplied by a plus 2. And then we're subtracting the value at 0, right? So we're subtracting -2 to the power of 0, multiplied by 0 + 2. Let's simplify We're going to get -2 E to the power of negative A divided by 2 multiplied by A + 2. Plus the power of 0 is 1, so we're adding 2 multiplied by 2, right? In other words, we're adding 4. And now what we can do is simply write this expression. In terms of a fraction. For the first term, notice that we get negative 2 multiplied by A + 2. Divided by E to the power of a divided by 2, since we have that negative exponent, we're adding 4, right? So we can find the common denominator which is E to the power of A divided by 2. Which essentially means that in the numerator, we can add 4 E to the power of a divided by 2. And now we can also factor out too. Let's go ahead and write equals, factoring out to, we get 2, and C's. We're going to get E to the power of a divided by 2 minus A minus 2, right? Which is divided by E to the power of a divided by 2. That's our final answer and thank you for watching.

82. A family of exponentials The curves y = x * e^(-a * x) are shown in the figure for a = 1, 2, and 3.
c. Find the area of the region bounded by y = x * e^(-a * x) and the x-axis on the interval [0, b]. Because this area depends on a and b, we call it A(a, b).

Key Concepts

Definite Integral and Area Under a Curve

Integration of Exponential Functions

Parameter Dependence in Families of Functions

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