# How do I evaluate this integral, given a table of values for f(x) and f'(x)?

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:

• It would help if you used the programming language that is designed to give a better picture of your writing so that it is more readable on a mathematical level rather than having to read lines of text. May 8, 2016 at 22:15

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 (https://en.wikipedia.org/wiki/Integration_by_substitution#Definition) 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.

• Okay what you're doing makes sense for sure, but I know that the answer is -1/2, and if we substitute the table values into your answer, we get -3/2. If it was instead f(-1) - f'(1) we would get the answer to be -1/2, but I can't see where you made any mistakes, so I'm not sure how they got that answer May 8, 2016 at 23:20

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 $3/2$.

• You substituted the whole $u = f(\cos(2x))$ and still got the same answer as me (except for the type in the last line, missing $-$ sign). I'm starting to believe that the professor's answer is wrong. May 9, 2016 at 8:37

Hermite interpolation finds the polynomial that fits a function given function and derivative values.

https://en.wikipedia.org/wiki/Hermite_interpolation

Construct the Hermite interpolation polynomial for your data and then integrate it from min to max value.