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

 To evaluate the integral \(\int_0^2 x f(x) \, dx\), we need to consider the piecewise nature of the function \(f(x)\) on the intervals \([0, 1]\) and \([1, 2]\). 

1. Interval \([0, 1]\): For \(x \in [0, 1]\), the function is defined as: \(f(x) = e^{\min(x^2, x-[x])}\). Since \(x - [x] = x\) in this interval (as \([x] = 0\) for \(0 \leq x < 1\)), we have: \(\min(x^2, x) = x^2\) when \(x \in [0, 1]\). Thus, \(f(x) = e^{x^2}\). Therefore, the integral becomes: \(\int_0^1 x e^{x^2} \, dx\).
2. To solve the integral \(\int_0^1 x e^{x^2} \, dx\), we use the substitution method by letting \(u = x^2\), so \(du = 2x \, dx\) or \(x \, dx = \frac{1}{2} du\). The limits change from \(x = 0 \to u = 0\) to \(x = 1 \to u = 1\). Thus, the integral becomes: \(\int_0^1 \frac{1}{2} e^u \, du = \frac{1}{2} [ e^u ]_0^1 = \frac{1}{2} (e - 1)\).
3. Interval \([1, 2]\): For \(x \in [1, 2]\), the function is defined as: \(f(x) = e^{[x - \log_ex]}\). Since \([x] = 1\) for \(1 \leq x < 2\), we have: \(f(x) = e^{1}\). Thus, the integral becomes: \(\int_1^2 x \cdot e \, dx = e \int_1^2 x \, dx\).
4. To solve the integral \(\int_1^2 x \, dx\), we calculate: \(\int_1^2 x \, dx = \left[ \frac{x^2}{2} \right]_1^2 = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}\). Therefore, \(\int_1^2 x \cdot e \, dx = e \cdot \frac{3}{2} = \frac{3e}{2}\).
5. Adding both integrals, we get: \(\int_0^2 x f(x) \, dx = \frac{1}{2}(e - 1) + \frac{3e}{2}\).
6. Simplifying, we find: \(\frac{1}{2} e - \frac{1}{2} + \frac{3e}{2} = 2e - \frac{1}{2}\).

Thus, the value of the integral \(\int_0^2 x f(x) \, dx\) is \(\boxed{2e - \frac{1}{2}}\).