Find the exact area of the shaded region, where f(x)=1/2xex 

The Correct Answer and Explanation is:

To find the exact area of the shaded region between the curves f(x)=12xexf(x) = \frac{1}{2}xe^x and g(x)=x2ln⁡xg(x) = x^2 \ln x over the interval x=1x = 1 to x=2x = 2, we evaluate the definite integral of their difference:

Area=∫12(12xex−x2ln⁡x)dx\text{Area} = \int_{1}^{2} \left( \frac{1}{2}xe^x – x^2 \ln x \right) dx

We start with the integral of 12xex\frac{1}{2}xe^x. Using integration by parts:

Let u=xu = x, dv=exdxdv = e^x dx, so du=dxdu = dx and v=exv = e^x. Then,

∫xexdx=xex−∫exdx=xex−ex\int xe^x dx = xe^x – \int e^x dx = xe^x – e^x

Thus,

∫12xexdx=12(xex−ex)=12xex−12ex\int \frac{1}{2}xe^x dx = \frac{1}{2}(xe^x – e^x) = \frac{1}{2}xe^x – \frac{1}{2}e^x

Evaluating from 1 to 2:

[12xex−12ex]12=(e2−12e2)−(12e−12e)=12e2\left[\frac{1}{2}xe^x – \frac{1}{2}e^x\right]_1^2 = \left(e^2 – \frac{1}{2}e^2\right) – \left(\frac{1}{2}e – \frac{1}{2}e\right) = \frac{1}{2}e^2

Now compute the integral of x2ln⁡xx^2 \ln x. Use integration by parts again:

Let u=ln⁡xu = \ln x, dv=x2dxdv = x^2 dx, then du=1xdxdu = \frac{1}{x} dx, v=x33v = \frac{x^3}{3}

∫x2ln⁡xdx=x33ln⁡x−∫x33⋅1xdx=x33ln⁡x−13∫x2dx=x33ln⁡x−x39\int x^2 \ln x dx = \frac{x^3}{3} \ln x – \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x – \frac{1}{3} \int x^2 dx = \frac{x^3}{3} \ln x – \frac{x^3}{9}

Evaluate from 1 to 2:

[x33ln⁡x−x39]12=(83ln⁡2−89)−(0−19)=83ln⁡2−79\left[\frac{x^3}{3} \ln x – \frac{x^3}{9}\right]_1^2 = \left(\frac{8}{3}\ln 2 – \frac{8}{9}\right) – (0 – \frac{1}{9}) = \frac{8}{3}\ln 2 – \frac{7}{9}

So the exact area is:

12e2−(83ln⁡2−79)=12e2−83ln⁡2+79\frac{1}{2}e^2 – \left(\frac{8}{3} \ln 2 – \frac{7}{9}\right) = \frac{1}{2}e^2 – \frac{8}{3}\ln 2 + \frac{7}{9}

Correct Answer:

79+12e2−83ln⁡2\boxed{\frac{7}{9} + \frac{1}{2}e^2 – \frac{8}{3} \ln 2}