Question

Find the area between y=4sinx and y=10cosx over the interval [0,π]. Sketch the curves if necessary.

Answer 1

$\sin x\ge0$ for all $x\in[0,\pi]$, while $\cos x>0$ for $0$ and $\cos x$ for $\dfrac\pi2$. So you already know that $4\sin x>10\cos x$ over the second half of the interval.

In the first half, $4\sin x$ is an increasing function, from $4\sin0=0$ to $4\sin\dfrac\pi2=4$. Meanwhile $10\cos x$ is a decreasing function, from $10\cos0=10$ to $10\cos\dfrac\pi2=0$. Therefore there must be a point between 0 and $\dfrac\pi2$ where the two curves intersect. So we find it:

$4\sin x=10\cos x\implies \tan x=\dfrac{10}4=\dfrac52\implies x=\arctan\dfrac52\approx1.1903$

So we know that $10\cos x>4\sin x$ in $\left[0,\arctan\dfrac52\right)$, and $4\sin x>10\cos x$ in $\left(\arctan\dfrac52,\pi\right]$.

The area between the curves is then

$\displaystyle\int_0^{\arctan(5/2)}(10\cos x-4\sin x)\,\mathrm dx+\int_{\arctan(5/2)}^\pi(4\sin x-10\cos x)\,\mathrm dx=4\sqrt{29}$