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 (Happy)

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- May 10th 2011, 08:47 PMTim82Gradient of a curve
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 (Happy) - May 10th 2011, 09:00 PMProve It
If you're finding f(pi), just substitute x = pi and simplify...

- May 10th 2011, 09:49 PMislam
- May 11th 2011, 04:15 AMHallsofIvy
"f1" is not a standard notation for the ordinary derivative but you may indeed be right.

Tim82, how about clarifying? - May 11th 2011, 04:27 AMSambit
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 $\displaystyle f(\pi)=\frac{{\pi}^2}{2}-\frac{\cos 2\pi}{2}+\frac{3}{2}=\frac{{\pi}^2}{2}-\frac{1}{2}+\frac{3}{2}=\frac{{\pi}^2}{2}+1$