Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Solution 1

Following the hint, we want to find a straight line that is below $y = \sin (x)$ with minimal error.

To do this, we require the straight line to pass through $y = \sin (x)$ at both $x = 0$ and $x = π / 2$ , which are the endpoints of the domain of integration.

Since $y = \sin (0) = 0$ at $x = 0$ , we look for a straight line that passes through the origin with formula $y = f (x) = c x$ . Then we put $x = π / 2$ , $y = \sin (π / 2) = 1$ to find $1 = c π / 2$ , so $c = 2 / π$ .

Note that the slope of the line $c = 2 / π < 1$ , while the initial slope of $y = \sin (x)$ is given by $\cos (0) = 1$ , so the linear function $y = (2 / π) x$ is really below $y = \sin (x)$ in the interval $[0, π / 2] .$ Hence, we can deduce the following inequalities:

\begin{aligned} \frac{2}{π} x & \leq \sin x \\ \Rightarrow - a \sin x & \leq - \frac{2 a}{π} x \\ \Rightarrow e^{- a \sin x} & \leq e^{- \frac{2 a}{π} x} \end{aligned}

because $a > 0$ and $e^{x}$ is a strictly increasing function. So by comparison, we obtain

\begin{aligned} \int_{0}^{π / 2} e^{- a \sin x} d x & \leq \int_{0}^{π / 2} e^{- \frac{2 a}{π} x} d x \\ = - \frac{π}{2 a} {[e^{- \frac{2 a}{π} x}]}_{0}^{π / 2} \\ = - \frac{π}{2 a} (e^{- a} - 1) \\ = \frac{π}{2 a} (1 - e^{- a}) < \frac{π}{2 a} \end{aligned}

because $e^{- a} > 0$ for any $a$ .