The area of the (right) triangle AOB is AO*OB/2. We have to find the equation of normal and the points A and B, its x- and y- intercepts.
y(x) = 4x + 3cos(2x).
Denote x0 = pi/4, y0 = y(x0) = pi + 3cos(pi/2) = pi.
Now find the equation of the normal.
1. It passes the point (x0, y0).
2. The slope of the tangent line at the point (x0, y0) is y'(x0).
y'(x) = 4 - 6sin(2x), y'(x0) = 4 - 6sin(pi/2) = -2.
So the slope of the normal is -1/(-2) = 1/2 and the equation of the normal is
y-y0 = (1/2)*(x-x0), or y-pi = (1/2)(x-pi/4).
For the point A put y=0 and obtain x = 2*(-pi) + pi/4 = -(7/4)pi. So AO = (7/4)pi.
For the point B put x=0 and obtain y = pi - pi/8 = (7/8)pi. So OB = (7/8)pi.
At last, the area = (1/2)*(7/4)pi*(7/8)pi = (49/64)pi^2. It is approx.= 7.556.
please ask if anything is unclear