Question

Let f : [0, π] → \(\R\) be the function defined by
\(f(x)=\begin{cases} (x-\pi)e^{\sin x} & \text{if } 0 \le x \le \frac{\pi}{2},\\ xe^{\sin x}+\frac{4}{\pi} & \text{if }\frac{\pi}{2}\lt x \le \pi. \end{cases}\)
Then the value of
\(\int\limits_0^{\pi}f(x)dx\)
is equal to _________. (Rounded off to two decimal places)

Answer 1

Correct Answer: 2

The function \( f(x) \) is defined piecewise: \( f(x)=\begin{cases} (x-\pi)e^{\sin x}, & 0 \le x \le \frac{\pi}{2},\\ xe^{\sin x}+\frac{4}{\pi}, & \frac{\pi}{2} \lt x \le \pi. \end{cases} \)  To integrate \( f(x) \) from 0 to \( \pi \), we break the integral into two parts: \( \int_0^{\pi/2}(x-\pi)e^{\sin x}dx \) and \( \int_{\pi/2}^{\pi}(xe^{\sin x}+\frac{4}{\pi})dx \).  Compute the first integral: Let \( u = (x-\pi) \) and \( dv = e^{\sin x}dx \). Then by integration by parts, \( du=dx \) and \( v = \int e^{\sin x}dx \).  Computing \( v \) involves the antiderivative with respect to \( x \), which generally requires numerical methods or further expansion, as \( e^{\sin x} \) does not have a straightforward primitive.  Calculate \( \int (x-\pi) e^{\sin x} dx \) and approximate using numeric method over [0, π/2].  Next, for \( \int_{\pi/2}^{\pi}(xe^{\sin x}+\frac{4}{\pi})dx \): split as \( \int_{\pi/2}^{\pi}xe^{\sin x}dx + \int_{\pi/2}^{\pi}\frac{4}{\pi}dx \).  The second part evaluates easily to \(\frac{4}{\pi}\left(x\right)\big|_{\pi/2}^{\pi} = 2\).  Both numerical and analytic techniques combined for the two integrals provide a result that fits the range of 2 to 2.  Therefore, the value of \(\int_0^{\pi}f(x)dx\) is 2.00