The problem gives me a list of values for $f(x)$ and $f'(x)$. I tried solving the problem using integration by parts, allowing $u=\sin(2x), u'=2\cos(2x), v'=f'(\cos(2x))$, and $v=f(\cos(2x))$. I use parts again and come up with $v'=f(\cos(2x))$ and I think this is where I'm stuck because I'm not sure how to integrate that.. Can I just turn it into $v=\cos(2x)$?? Am I even taking the correct approach using parts?
I know that when solving an integration problem just using functions of $x$ and values, that you can construct a riemann sum and solve it that way, but the incorporation of $f(x)$ and $f'(x)$ is throwing me off.
Here is a picture of the problem:
3 Answers
$\begingroup$I totally don't know if this is right what I'm doing, but I'd like to know if my approach is correct.
If we straightly substitute in $$u = \cos(2x)$$ We'd have
$$du = -2\sin(2x)\; dx $$ Or equivalently $$-\frac{1}{2}\; \text{d}u = \sin(2x)\; \text{d}x $$
When $x=\pi / 2$, then $u = \cos(2\pi/2) = \cos(\pi) = -1$ and when $x = 0$, $u = \cos(0) =1$.
So we can transform the whole integral from
$$\int_{0}^{\pi/2} \sin(2x) \cdot f'(\cos(2x)) \; \text{d}x$$ to $$\int_{1}^{-1} -\frac{1}{2}\cdot f'(u) \; \text{d}u$$ And I guess it will help that we know $f(1)$, $f(-1)$, $f'(-1)$ and $f'(1)$.
Since then we have: $$\int_{1}^{-1} -\frac{1}{2}\cdot f'(u) \; \text{d}u = -\frac{1}{2} \int_1^{-1} f'(u) \; \text{d}u = -\frac{1}{2}\left(f(-1) - f(1) \right) $$ And we can compute that constant now because we know the function values.
However it seems awfully suspicious that this answer doesn't need any function values of $f'()$ at all.. Hm..
EDIT: Weird, even the formula on wikipedia () straightly says
$$\int_{\varphi(a)}^{\varphi(b)} f(x) \; \text{d}x = \int_a^b f(\varphi(t))\varphi'(t) \; \text{d}t$$
Working from the right side we get from the original integral $$\int_{0}^{\pi/2} \sin(2x) \cdot f'(\cos(2x)) \; \text{d}x = -\frac{1}{2}\int_{0}^{\pi/2} -2\sin(2x) \cdot f'(\cos(2x)) \; \text{d}x$$ We can see that in this case $$\varphi(x) = \cos(2x)$$ $$\varphi'(x) = -2\sin(2x)$$ $$a = 0 \quad b=\pi/2$$ And therefore the bounds on the left integral must be $$\varphi(a) = \cos(0) = 1$$ $$\varphi(b) = \cos(2\cdot \pi/2) = -1$$ And now we get the left side as
$$\int_1^{-1} -\frac{1}{2}f'(u) \; \text{d}u = -\frac{1}{2}\left(f(-1) - f(1)\right) = -\frac{1}{2}(10-7) = -\frac{3}{2}$$
Which is the same as before -.-.
If the upper bound was however $0$, we'd have $$\int_1^{0} -\frac{1}{2}f'(u) \; \text{d}u = -\frac{1}{2}\left(f(0) - f(1)\right) = -\frac{1}{2}(8-7) = -\frac{1}{2}$$ Nope, no idea what's wrong. Now I'd also like to learn what the correct answer is.
$\endgroup$ 1 $\begingroup$Take $f(cos(2x))=t$.
Then we have $f'(cos(2x))*-sin(2x)*2 dx= dt$
Substitute this into the integral to get $\int -1/2 dt$
The limits will be $f(1)=7$ to $f(-1)=10$, which gives you
$\int_{7}^{10} -1/2 dt = -3/2$
So the required answer is $3/2$.
$\endgroup$ 1 $\begingroup$Hermite interpolation finds the polynomial that fits a function given function and derivative values.
Construct the Hermite interpolation polynomial for your data and then integrate it from min to max value.
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