integration – NST First Year Maths

2020 Paper 1 Question 6

25th Jan 2022nstmathsupervisorLeave a comment

$I = \int x \sqrt{3-2\,x}\, dx$

You can do this by substitution.

Let $u = 3 -2x$

$\Rightarrow dx = -\frac{1}{2} du$ and $x = \frac{1}{2} \left( 3-u \right )$

$\begin{aligned} \therefore I &= \int \frac{(3-u)}{2} \, u^{1/2} \, \left(-\frac{1}{2} \right) \,du \\ &= -\frac{1}{4} \int \left( 3u^{1/2} - u^{3/2} \right) \, du \\ &= -\frac{1}{4} \left( 2u^{3/2}-\frac{2}{5} u^{5/2} \right) + C \\ &= -\frac{1}{2} u^{3/2} \left( \frac{5-u}{5} \right) + C \\ &= -\frac{1}{2} (3-2x)^{3/2} \left( \frac{2+2x}{5} \right) + C \\ &= -\frac{1}{5} (3-2x)^{3/2} \left( 1+x \right) + C \end{aligned}$

This is probably the most compact form. Other forms are possible, for example by writing $(3-2x)^{3/2} = (3-2x)^{1/2}(3-2x)$ .

2013 Paper 2 Question 13

3rd Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Let $y = x/2$

$dy = dx/2$

Therefore,

$I = 2\,\int_0^{\infty} e^{-y^2}\,dy$

Now, given

$\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt \pi$

we have, due to the even integrand

$\int_{0}^{\infty}e^{-x^2}dx = \frac{1}{2}\sqrt \pi$

Thus,

$I = 2 \left( \frac{1}{2} \sqrt \pi \right) = \sqrt \pi$

(b)

(i)

The key is to rewrite the integrand and use integration by parts,

$I_n = \int_{-\infty}^{\infty} x^{n-1}\,x\,e^{-x^2}dx$

Now let $u =x^{n-1}$ and $dv/dx=x\,e^{-x^2}$

$du = (n-1)x^{n-2}\,dx$

and

$v = -\frac{1}{2}e^{-x^2}$

So

$I_n = \left[ -\frac{1}{2}x^{n-1}e^{-x^2} \right]_{-\infty}^{\infty} + \frac{1}{2} \int_{-\infty}^{\infty}(n-1) x^{n-2}e^{-x^2}\,dx$

The exponential dominates the polynomial, so the first term is zero, and the final integral belongs to the family of given integrals, so

$I_n = \frac{n-1}{2} \, I_{n-2}$