Question

Let the function f: [0,2] \(\rightarrow\) R be defined as \(f(x) =   \begin{cases}     e^{min}{x^2,x-[x]}       & \quad x \in[0,1]\\     e^{[x-log_ex]}  & \quad x\in[1,2]   \end{cases}\)
where [t] denotes the greatest integer less than or equal to t. Then the value of the integral \(\int^2_0xf(x)dx\) is 

• A. \(\bigg(e^2+\frac{1}{2}\bigg)\)
• B. 2e-1
• C. \(1+\frac{3e}{2}\)
• D. \(2e-\frac{1}{2}\)

Answer 1

The Correct Option is D

Step 1: Split the integral

We split the integral into two parts based on the definition of \(F(x)\):

\[ \int_0^2 xF(x)dx = \int_0^1 xe^{x^2}dx + \int_1^2 xe^x dx \]

Let:

• \(I_1 = \int_0^1 xe^{x^2}dx\)
• \(I_2 = \int_1^2 xe^x dx\)

Step 2: Solve \(I_1 = \int_0^1 xe^{x^2}dx\)

Use the substitution \(u = x^2 \Rightarrow du = 2x dx\). The limits change as:

• When \(x = 0, u = 0\)
• When \(x = 1, u = 1\)

The integral becomes:

\[ I_1 = \frac{1}{2} \int_0^1 e^u du \]

Evaluate the integral of \(e^u\):

\[ I_1 = \frac{1}{2} \left[e^u\right]_0^1 = \frac{1}{2} \left(e^1 - e^0\right) = \frac{1}{2}(e - 1) \]

Step 3: Solve \(I_2 = \int_1^2 xe^x dx\)

Since \(e^x\) is constant with respect to \(x\), the integral becomes:

\[ I_2 = e \int_1^2 x dx \]

Evaluate the integral of \(x\):

\[ \int_1^2 x dx = \frac{x^2}{2} \bigg|_1^2 = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]

Thus:

\[ I_2 = e \cdot \frac{3}{2} = \frac{3}{2}e \]

Step 4: Combine the results

Add \(I_1\) and \(I_2\) to find the total integral:

\[ \int_0^2 xF(x)dx = I_1 + I_2 = \frac{1}{2}(e - 1) + \frac{3}{2}e \]

Combine terms:

\[ \int_0^2 xF(x)dx = \frac{1}{2}e - \frac{1}{2} + \frac{3}{2}e = \frac{4}{2}e - \frac{1}{2} = 2e - \frac{1}{2} \]

Final Answer

\[ \int_0^2 xF(x)dx = 2e - \frac{1}{2} \]