1. ## Integration question

Let f(x) be a continous function such that its first two derivatives f'(x) and f''(x) are continous. The tangents to the graph of f(x) at the points with abscissa x=a and x=b make with the X-axies angles π / 3 and π /4 respectively. then the value of the integral

Integrate [ f'(x) * f''(x) dx ] from a to b equals?

a. 1-root3
b. 0
c. 1
d. -1

Thanks

2. Hello,
Originally Posted by champrock
Let f(x) be a continous function such that its first two derivatives f'(x) and f''(x) are continous. The tangents to the graph of f(x) at the points with abscissa x=a and x=b make with the X-axies angles π / 3 and π /4 respectively. then the value of the integral

Integrate [ f'(x) * f''(x) dx ] from a to b equals?

a. 1-root3
b. 0
c. 1
d. -1

Thanks
I don't know how to solve the final stuff (I guess you have notes about values of the derivative if the tangent makes a certain angle with the x-axis), but I hope this can help you :

$\int_a^b f'(x) f''(x) ~dx=\frac 12 \int_a^b 2 f'(x) [f'(x)]' ~dx$

And this integrates into :
$\left.\frac 12 ~[f'(x)]^2 \right|_a^b$

3. no notes here.

What I had initially attempted was take f'(x) = t so f''(x) dx = dt therefore the integral reduces to Integrate[t d(t)]

But dont know exactly how to interprete (a,b) here.

4. Originally Posted by champrock
no notes here.

What I had initially attempted was take f'(x) = t so f''(x) dx = dt therefore the integral reduces to Integrate[t d(t)]

But dont know exactly how to interprete (a,b) here.
But please next time, write down what you've tried;...

$\int_a^b t ~dt=\left. \frac 12 t^2 \right|_a^b=\left. \frac 12 [f'(x)]^2 \right|_a^b=\frac 12 [f'(b)]^2-\frac 12 [f'(a)]^2$

Now you know that the tangent to the curve at point a, has a slope of f'(a).
If it forms an angle of pi/3 with the x axis, it means that its slope is tan(pi/3) ---> f'(a)=tan(pi/3) and f'(b)=tan(pi/4)

There is an example here : Vectors Aligned with X-Y Axes, but I can't find a good link (and it's time to eat ^^)

5. actually, I was not at all sure whether this was the right way or not, so I didnt bother to write it. Will keep it in mind.

Enjoy! and thanks once again for boththe questions!