Question

If $f(x)$ is a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ is a function that satisfies $f(x)+g(x)=x^{2}$. Then the value of the integral $\int_{0}^{1}f(x)g(x)dx$ is

• A. $e-\frac{e^{2{2}-\frac{5}{2}$
• B. $e+\frac{e^{2{2}-\frac{3}{2}$
• C. $e-\frac{e^{2{2}-\frac{3}{2}$
• D. $e+\frac{e^{2{2}+\frac{5}{2}$

Hint:

Logic Tip: The integral $\int x^n e^x dx$ follows a simple pattern: $e^x [x^n - nx^{n-1} + n(n-1)x^{n-2} - .......]$. For $n=2$, it's simply $e^x(x^2 - 2x + 2)$. Memorizing this pattern drastically speeds up integration by parts.

Answer 1

The Correct Option is C

Concept:
First, solve the simple differential equation $f'(x) = f(x)$ to find the explicit form of $f(x)$. Then, use the given relation to find $g(x)$. Finally, calculate the definite integral of their product using integration by parts.
Step 1: Find the function f(x).
We are given the differential equation: $$f'(x) = f(x)$$ $$\frac{f'(x)}{f(x)} = 1$$ Integrate both sides with respect to $x$: $$\int \frac{f'(x)}{f(x)} dx = \int 1 \, dx$$ $$\ln|f(x)| = x + C$$ Using the initial condition $f(0) = 1$: $$\ln|1| = 0 + C \implies 0 = C \implies C = 0$$ Thus, $\ln|f(x)| = x \implies f(x) = e^x$.
Step 2: Find the function g(x) and set up the integral.
We are given the relationship: $$f(x) + g(x) = x^2$$ Substitute $f(x) = e^x$: $$e^x + g(x) = x^2 \implies g(x) = x^2 - e^x$$ Now, substitute $f(x)$ and $g(x)$ into the required integral: $$I = \int_{0}^{1} f(x)g(x) dx = \int_{0}^{1} e^x(x^2 - e^x) dx$$ $$I = \int_{0}^{1} (x^2 e^x - e^{2x}) dx$$
Step 3: Evaluate the integral.
Split the integral into two parts: $$I = \int_{0}^{1} x^2 e^x dx - \int_{0}^{1} e^{2x} dx$$ For the first part, use integration by parts twice (or the tabular method): $$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$$ $$= x^2 e^x - 2(x e^x - \int e^x dx) = x^2 e^x - 2x e^x + 2e^x = e^x(x^2 - 2x + 2)$$ Now evaluate both parts from 0 to 1: $$I = [e^x(x^2 - 2x + 2)]_0^1 - \left[\frac{e^{2x{2}\right]_0^1$$ $$I = \left( e^1(1 - 2 + 2) - e^0(0 - 0 + 2) \right) - \left( \frac{e^2}{2} - \frac{e^0}{2} \right)$$ $$I = (e(1) - 1(2)) - \left( \frac{e^2}{2} - \frac{1}{2} \right)$$ $$I = e - 2 - \frac{e^2}{2} + \frac{1}{2}$$ $$I = e - \frac{e^2}{2} - \frac{3}{2}$$