Question

Let \(f : \mathbb{R}\to\mathbb{R}\) be defined as \(f(x)=e^{-x}\sin x\). If \(F:[0,1]\to\mathbb{R}\) is a differentiable function with that \(F(x)=\int_0^x f(t)\,dt\), then the value of \(\int_0^1 \left(F'(x)+f(x)\right)e^{x}\,dx\) lies in the interval

• A. \(\left[\frac{331}{360},\frac{334}{360}\right]\)
• B. \(\left[\frac{330}{360},\frac{331}{360}\right]\)
• C. \(\left[\frac{335}{360},\frac{336}{360}\right]\)
• D. \(\left[\frac{327}{360},\frac{329}{360}\right]\)

Hint:

By the fundamental theorem, \(F'(x)=f(x)\), so the integral is \(2\int_0^1\sin x\,dx=2(1-\cos 1)\). Expand \(\cos 1\) as an alternating series to compare with \(\frac{331}{360}\).

Answer 1

The Correct Option is B

Since \(F(x)=\int_0^x f(t)\,dt\), \(F'(x)=f(x)\).
So \(\int_0^1 (F'(x)+f(x))e^x dx=\int_0^1 2e^{-x}\sin x\cdot e^{x}dx=2\int_0^1\sin x\,dx=2(1-\cos 1)\).
\(\cos 1=1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\dots\), so \(2(1-\cos 1)=1-\frac{1}{12}+\frac{1}{360}-\frac{1}{20160}+\dots=\frac{331}{360}-\frac{1}{20160}+\frac{1}{1814400}-\dots\)
The series alternates with decreasing terms, hence \(\frac{331}{360}-\frac{1}{20160}\lt 2(1-\cos 1)\lt\frac{331}{360}\), and the value is clearly greater than \(\frac{330}{360}\) (numerically about 0.91940, while \(\frac{331}{360}\approx 0.91944\) and \(\frac{330}{360}\approx 0.91667\)).
So the value lies in \(\left[\frac{330}{360},\frac{331}{360}\right]\).
Answer: (2) \(\left[\frac{330}{360},\frac{331}{360}\right]\)