Why does integration by parts work?

Let's consider the form of the formula we use for integration by parts: the integral of u * dv/dx = uv - the integral of v * du/dx. We know that integration is the inverse of differentiation, so we should be able to differentiate both sides to get back to u * dv/dx. The left hand side of this equation obviously satisfies this. d/dx(uv - integral of v * du/dx) = d/dx(uv) - d/dx(integral of v * du/dx).
For d/dx(uv), we'll use the product rule: d/dx(uv) = du/dx * v + u * dv/dx. For d/dx(integral of v * du/dx), we'll just use the fact that integration is the inverse of differentiation, d/dx(integral of v * du/dx) = v * du/dx. So, d/dx(uv) - d/dx(integral of v * du/dx) = (du/dx * v + u * dv/dx) - (du/dx * v) = u * dv/dx. Since the derivative of uv - integral of v * du/dx = u * dv/dx, the integral of u * dv/dx = uv - integral of v * du/dx.

TS

Related Maths A Level answers

All answers ▸

A particle of mass 0.5 kg is moving down a rough slope (with coefficient of friction = 0.2) inclined at 30 degrees to the horizontal. Find the acceleration of the particle. Use g = 9.8 ms^-2.


Differentiate 2x^3+23x^2+3x+5 and find the values of x for which the function f(x) is at either at a maximum or minimum point. (Don't need to specify which is which)


Integrate the following function: f(x) = ln(x)


If f(x) = x^2 - 3x + 2, find f'(x) and f''(x)