Math Doubts

Proof of Integration by parts

The mathematical formula for the integration by parts can be derived in integral calculus by the concepts of differential calculus.

Product Rule of Differentiation

$f{(x)}$ and $g{(x)}$ are two functions in terms of $x$. As per the product rule of differentiation, the derivative of the product of the functions $f{(x)}$ and $g{(x)}$ can be written into its equivalent form.

$\dfrac{d}{dx}{\, \Big({f{(x)}}{g{(x)}}\Big)}$ $\,=\,$ $f{(x)}\dfrac{d}{dx}{\, {g{(x)}}}$ $+$ $g{(x)}\dfrac{d}{dx}{\, {f{(x)}}}$

Product Rule in Differential form

Take $u = f{(x)}$ and $v = g{(x)}$. Now, express the derivative product rule in differential form.

$\implies$ $\dfrac{d}{dx}{\, (uv)}$ $\,=\,$ $u\dfrac{d}{dx}{\, v}$ $+$ $v\dfrac{d}{dx}{\, u}$

$\implies$ $\dfrac{d(uv)}{dx}$ $\,=\,$ $u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}$

Multiply both sides of the equation by the differential element $dx$.

$\implies$ $\dfrac{d(uv)}{dx} \times dx$ $\,=\,$ $\Big(u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}\Big) \times dx$

$\implies$ $\dfrac{d(uv)}{dx} \times dx$ $\,=\,$ $u\dfrac{dv}{dx} \times dx$ $+$ $v\dfrac{du}{dx} \times dx$

$\implies$ $d(uv) \times \dfrac{dx}{dx}$ $\,=\,$ $udv \times \dfrac{dx}{dx}$ $+$ $vdu \times \dfrac{dx}{dx}$

$\implies$ $d(uv) \times 1$ $\,=\,$ $udv \times 1$ $+$ $vdu \times 1$

$\implies$ $d(uv)$ $\,=\,$ $udv$ $+$ $vdu$

Integrate the Differential equation

Now, integrate both sides of the differential equation.

$\implies$ $\displaystyle \int d(uv)$ $\,=\,$ $\displaystyle \int udv$ $+$ $\displaystyle \int vdu$

$\implies$ $uv$ $\,=\,$ $\displaystyle \int udv$ $+$ $\displaystyle \int vdu$

$\implies$ $uv$ $-$ $\displaystyle \int vdu$ $\,=\,$ $\displaystyle \int udv$

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int udv$ $\,=\,$ $uv$ $-$ $\displaystyle \int vdu$

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