Q:
The gradient of a curve with equation y=f(x) is given by f1 (x) = x + sin2x.
If f(0) = 1, find f(\pi )
Any help would be appreciated
Exactly. So $\displaystyle f(x)=\int x dx+\int \sin 2x dx=\frac{x^2}{2}-\frac{\cos 2x}{2}+c$
Putting $\displaystyle f(0)=1$ we have $\displaystyle c=\frac{3}{2}$
So $\displaystyle f(x)=\frac{x^2}{2}-\frac{\cos 2x}{2}+\frac{3}{2}$
So $\displaystyle f(\pi)=\frac{{\pi}^2}{2}-\frac{\cos 2\pi}{2}+\frac{3}{2}=\frac{{\pi}^2}{2}-\frac{1}{2}+\frac{3}{2}=\frac{{\pi}^2}{2}+1$