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<x<\frac\pi2` and
`\cos x<0` for
`\frac\pi2<x<\pi`. 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\frac\pi2=4`. Meanwhile
`10\cos x` is a decreasing function, from
`10\cos0=10` to
`10\cos\frac\pi2=0`. Therefore there must be a point between 0 and
`\frac\pi2` where the two curves intersect. So we find it:


`4\sin x=10\cos x\implies \tan x=\frac{10}4=\frac52\implies x=\arctan\frac52\approx1.1903`

So we know that
`10\cos x>4\sin x` in
`\left[0,\arctan\frac52\right)`, and
`4\sin x>10\cos x` in
`\left(\arctan\frac52,\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√(29)`