How do I solve the following definite integral using integration by parts?

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)?

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

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

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

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 :)

• 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 $$\left[u\int v dx\right]_a^b \ne u|_a^b \int_a^b v dx$$ ? Aug 2 at 17:45
• 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 $$u|_a^b\int_a^b vdx = (b-a)^2$$ Aug 2 at 19:57
• @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. Aug 3 at 0:22

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 $$G(c)=0$$.