3
$\begingroup$

Please refer to the diagram below. enter image description here

The diagram shows a curve with equation $y = cos {x\over 2} cos x$, for 0 $\le x$ $\le$ $\pi$, along with the $x$ and $y$-intercepts of the graph.

Question: By first finding $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$, explain why $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$ is smaller than $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$. You may refer to the graph provided for assistance.

I have found $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$ and have done so as shown: enter image description here

I then start to calculate $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$ as shown below.

enter image description here

Now, I realise I would also end up calculating $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$ eventually which implies that $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$ = $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$? But the question has already stated that $\vert \int_{\pi\over 3}^{2\pi\over 3} cos {x\over 2} cos x \,dx \vert$ will be smaller than $\int_{\pi\over 3}^{2\pi\over 3} \vert cos {x\over 2} cos x \vert\,dx$, so I'm not really sure how to proceed from here...

  • 0
    Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$. – 2017-01-02

1 Answers 1

0

Let $f(x)=\cos\frac{x}{2}\cos x$. Then $f$ has a change of sign at $x=\pi/2$, so $$ \int_{\pi/3}^{2\pi/3}|f(x)|\,dx = \int_{\pi/3}^{2\pi/3}|f(x)|\,dx = \int_{\pi/3}^{\pi/2}f(x)\,dx - \int_{\pi/2}^{2\pi/3}f(x)\,dx $$ On the other hand $$ \left|\int_{\pi/3}^{2\pi/3}|f(x)|\,dx\right| $$ does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have $$ \int_{\pi/3}^{2\pi/3}|f(x)|\,dx= (F(\pi/2)-F(\pi/3))-(F(2\pi/3)-F(\pi/2))= -F(2\pi/3)-F(\pi/2)+2F(\pi/2) $$ whereas $$ \left|\int_{\pi/3}^{2\pi/3}|f(x)|\,dx\right|= |F(2\pi/3)-F(\pi/3)| $$