Integration — common mistakes

In exams, if you are asked to 'show' or 'prove' a reduction formula, you must show all your working clearly, including your choice of u and dv/dx, and the evaluation of the $[uv]$ term. These derivation steps carry significant marks.

For arc length and surface area problems, the expression under the square root, such as $(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2$, is very often a perfect square. If your expression looks complicated, double-check your derivatives and then look for a way to factorise it into the form $(A+B)^2$.

It's a strategic choice. You want the new integral, $\int v \frac{du}{dx} \, dx$, to be simpler or related to the original integral. For integrals like $\int x^n f(x) \, dx$, choosing $u=x^n$ usually works because differentiating it reduces the power of x. For trigonometric integrals like $\int \sin^n x \, dx$, you split it into $\sin^{n-1} x \cdot \sin x$ and choose $u = \sin^{n-1} x$.

No, this is not a good strategy. Exam questions frequently ask you to 'Show that...' or 'Derive...' the formula itself. Marks are given for the derivation process. You must understand and be able to reproduce the derivation using integration by parts.

No, in this context, the terms 'mean value' and 'average value' of a function over an interval are used interchangeably. Both refer to the value $\frac{1}{b-a} \int_a^b f(x) \, dx$.

There are two likely culprits. First, check your derivatives ($rac{dy}{dx}$ or $rac{dx}{dt}, rac{dy}{dt}$) for simple errors. Second, and more commonly, check if the expression inside the square root simplifies to a perfect square. Exam questions are almost always designed so that this happens, leading to an integral you can solve.