2
$\begingroup$

How do I solve the following integral ($C_1$ and $C_2$ are constants)?

$$ I=\int_a^b e^{x\cdot(C_1-C_2)} \cdot g(x) dx $$

Question 1: I tried solving it using integration by parts with the product rule $\left(\int f(x)\cdot g(x) dx=f(x)\cdot \int g(x)dx-\int \frac{d}{dx}f(x)\cdot (\int g(x) dx) dx\right)$ and found zero as my result, but that is wrong because both of my functions ($e^{x\cdot(C_1-C_2)}$ and $g(x)$) have positive values within the specified interval. What's the equivalent formula for definite integration? Is the formula below correct?

$$ \int_a^b f(x)\cdot g(x) dx=(f(b)-f(a))\cdot \int_a^b g(x)dx-\int_a^b \frac{d}{dx}f(x)\cdot (\int_a^b g(x) dx) dx $$

Question 2: In the case of definite integration, can I treat the inner integral as a constant and bring it outside the main integral? In that case, we would have $$ \int_a^b \frac{d}{dx}f(x)\cdot \left(\int_a^b g(x) dx\right) dx=\left(\int_a^b g(x) dx\right)\cdot\int_a^b \frac{d}{dx}f(x) dx $$

Question 3: Is there any functional difference between these two integrals ($I_A$ and $I_B$) (the integration variable in the inner integral is denoted by different letters)?

$$ I_A=\int_a^b \frac{d}{dx}f(x)\cdot \left(\int_a^b g(x) dx\right) dx $$

$$ I_B=\int_a^b \frac{d}{dx}f(x)\cdot \left(\int_a^b g(s) ds\right) dx $$

$\endgroup$
2
  • $\begingroup$ "Is the formula below correct?" No. $\endgroup$
    – PNDas
    Aug 2 at 16:36
  • $\begingroup$ You may want to take a look at here $\endgroup$
    – by24
    Aug 2 at 16:38
3
$\begingroup$

The answer to 1 and 2 are the same: the formula you have used is incorrect. The correct formula is $$\int_a^b uv\,\textrm{d}x=\left[u\int v\,\textrm dx\right]_a^b - \int_a^b \Big(u'\cdot\int v\,\textrm dx\Big)\,\textrm dx $$ ($u $ and $v $ refer to functions of $x $)

In fact, this will make sense if you think of the Fundamental Theorem of Calculus.

The answer to 3 is yes. Because the integral only depends on the value of antiderivative at the two end points, the variable name doesn't matter. (It is called a dummy variable.) So, if you see an integral like you mentioned in 2 (though its not valid here), you may treat it as a constant.

Hope this helps. Ask anything if not clear :)

$\endgroup$
3
  • $\begingroup$ Thanks for the explanation! Just to be clear, in your explanation you included an indefinite integral inside the definite integral. Is that correct or is it supposed to be a definite integral as well? I'm asking because if it is supposed to be a definite integral in [a,b], then the rule I wrote in question 2 can be used, right? A second question: can you show me why $$ \left[u\int v dx\right]_a^b \ne u|_a^b \int_a^b v dx $$ ? $\endgroup$ Aug 2 at 17:45
  • 1
    $\begingroup$ just take a simple example $u=x, v=1$ you will find that $$\left[u\int vdx\right]_a^b = b^2 - a^2$$ and $$u|_a^b\int_a^b vdx = (b-a)^2$$ $\endgroup$ Aug 2 at 19:57
  • 1
    $\begingroup$ @williantafsilva: No, it is an indefinite integral inside. Why the result you stated is not true, you may think of which by using the Fundamental Theorem of Calculus. $$\int_a^bf(x)\,\textrm dx=F(b)-F(a)$$($F$ is antiderivative of $f$). If you let $f(x)=uv$ and put the antiderivative $F$ (by regular by parts) then you will observe that the limits will only be put outside of the two parts in the formula. $\endgroup$ Aug 3 at 0:22
2
$\begingroup$

The principal mistake you are making here is the fact that you don't integrate the function $g()$ before using it.

In fact you should have $$\int_{a}^b f(x).g(x)dx = \left[f(x).G(x)\right]_{a}^b - \int_{a}^b f'(x).G(x)dx\,,$$ where we suppose $G(x)=\int_{c}^x g(s)ds$ and $G(c)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.