First slide

Evaluation of definite integrals

Question

The least value of the function $F (x) = \int_{0}^{x}$ $(3 \sin u + 4 \cos u) d u$
on the interval $(5 π / 4, 4 π / 3]$ is

Moderate

A

$3 / 2 - \sqrt{3} / 2$
B

$\frac{5 - 4 \sqrt{3}}{2}$
C

$\frac{7 - 4 \sqrt{3}}{2}$
D

$\frac{9 - 4 \sqrt{3}}{2}$

Solution

We have $F' (x) = 3 \sin x + 4 \cos x .$ Since sin $x$ and cos $x$ assume negative values in the third quadrant,
we have $F^{'} (x) < 0$ for all $x \in (5 π / 4, 4 π / 3)$ so $F (x)$ assumes the least value at the point $x = 4 π / 3 .$
Thus the least value is $F (4 π / 3) = \int_{0}^{4 π / 3} (3 \sin u + 4 \cos u) d u$
$\begin{array}{l} = {(- 3 \cos u + 4 \sin u)|}_{0}^{4 π / 3} \\ = - 3 \cos \frac{4 π}{3} + 4 \sin \frac{4 π}{3} - (- 3) \\ = \frac{9}{2} - \frac{4 \sqrt{3}}{2} = \frac{9 - 4 \sqrt{3}}{2} . \end{array}$

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