Cfrgtkky

I'm trying to obtain the area between the curve of these two functions (for $x>0$), lets call them $f(x)=2cos(x)$ and $g(x)=x/2$ and my idea is to get the area under the curve of $f(x)$, then subtract the sum of these: the area under the curve of $g(x)$ [$0$, intersection point] and $f(x)$ [intersection point, $pi/2$]

Is this the right way or there's an easier way?

Thanks.

I guess you could use Wolfram's approximation for the intersection point: $xapprox1.25235$.

Then you get $approxint_0^{1.25235} (2cos x-frac x2)operatorname dx=[2sin x-frac {x^2}4]_0^{1.25235}=2sin 1.25235-frac{(1.25235)^2}4approx1.50735$ for the area between the curves.

I'm not sure what you mean by subtracting the sum of what you listed, but you just need to find find the area from x=0 to the intersection point. Like what someone else mentioned, it's approximately x=1.252. Then you just take the definite integral bounded by 0 on the left and 1.252 on the right, for both equations. Subtract g(x) from f(x) and you'll have your answer.