Definite integral integration by parts

Question

Can we write the integration by parts for definite integral the following way:

$$\int^a_b f(x)g(x)dx=f(x)\int^a_b g(x)dx-\int^a_b \left[ \dfrac{df(x)}{dx}\int^a_b g(x)dx \right]dx $$

My book gives the following formula for definite integral integration by parts:

$$\int^a_b f(x)g(x)dx=\left[f(x)\int g(x)dx\right]^a_b -\int^a_b \left[ \dfrac{df(x)}{dx}\int g(x)dx \right]dx $$

Are the two formulas equivalent or not? Why/Why not?

Answer 1

The equality$$\int^a_b f(x)g(x)dx=f(x)\int^a_b g(x)dx-\int^a_b \left[ \dfrac{df(x)}{dx}\int^a_b g(x)dx \right]dx $$cannot possibly be true, because $\displaystyle\int^a_b f(x)g(x)dx$ and $\displaystyle\int^a_b \left[ \dfrac{df(x)}{dx}\int^a_b g(x)dx \right]dx$ are numbers, whereas $\displaystyle f(x)\int^a_b g(x)dx$ depends upon $x$.

Answer 2

Integration by parts is defined by $$\int f(x) \, g(x) \, dx = f(x) \int g(u) \, du - \int f'(t) \left(\int^{t} g(u) \, du \right) \, dt.$$ When applying limits on the integrals they follow the form $$\int_{a}^{b} f(x) \, g(x) \, dx = \left[f(x) \int g(u) \, du\right]_{a}^{b} - \int_{a}^{b} f'(t) \left(\int^{t} g(u) \, du \right) \, dt.$$

Now, if $$\int_{a}^{b} f(x) \, g(x) \, dx = f(x) \int_{a}^{b} g(u) \, du - \int_{a}^{b} f'(t) \left(\int_{a}^{b} g(u) \, du \right) \, dt$$ then what has been descried is $$\int_{a}^{b} g(u) \, du$$ is a constant, say $c_{1}$ for which the remaining integration becomes $$\int_{a}^{b} f(x) \, g(x) \, dx = c_{1} \, f(x) - c_{1} \, \int_{a}^{b} f'(t) \, dt.$$

These resulting integrals are not the same, in any sense, unless $g(x)$ is a constant to begin with.

As a demonstration consider $f(x) = x, g(x) =1$ for which the proper way yields \begin{align} \int_{1}^{2} f(x) \, g(x) \, dx &= \left[ x \cdot x \right]_{1}^{2} - \int_{1}^{2} 1 \cdot x \, dx \\ &= (4 - 1) - \left[ \frac{x^2}{2}\right]_{1}^{2} = \frac{3}{2}. \end{align} The questioned method leads to \begin{align} \int_{1}^{2} f(x) \, g(x) \, dx &= x \int_{1}^{2} du - \int_{1}^{2} 1 \cdot \left(\int_{1}^{2} du \right) \, dt \\ &= x - \int_{1}^{2} dt \\ &= x - 1. \end{align}

From this example even if it had been asked for $$\int_{a}^{b} f(x) \, g(x) \, dx = \left[f(x) \int_{a}^{b} g(u) \, du\right]_{a}^{b} - \int_{a}^{b} f'(t) \left(\int_{a}^{b} g(u) \, du \right) \, dt$$ then the exampled result would be \begin{align} \int_{1}^{2} f(x) \, g(x) \, dx &= \left[x \int_{1}^{2} du\right]_{1}^{2} - \int_{1}^{2} 1 \cdot \left(\int_{1}^{2} du \right) \, dt \\ &= [x]_{1}^{2} - \int_{1}^{2} dt \\ &= 1 - 1 = 0. \end{align} which still leads to an incorrect result.