Recent content by Prove It

Why wouldn't you use partial fractions though?

I like your idea of sketching the graphs. Maybe start by working out the intercepts with the co-ordinate axes. E.g. for x + y + z = 2, the x intercept is x + 0 + 0 = 2 => x = 2. With the three intercepts it's pretty easy to draw the plane.

What did you have for part (i)?

Have you tried anything?

It might be worth noting that the integrand is an even function...

It's well known that $\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \sqrt{\pi} \end{align*}$, and due to the evenness of this function, that means $\displaystyle \begin{align*} \int_0^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \frac{\sqrt{\pi}}{2}...

$\displaystyle \begin{align*} \lim_{x \to \infty} \left( \frac{\mathrm{e}^{3\,x} - \mathrm{e}^{-3\,x}}{\mathrm{e}^{3x} + \mathrm{e}^{-3\,x}} \right) &= \lim_{x \to \infty} \left[ \left( \frac{\mathrm{e}^{3\,x} - \mathrm{e}^{-3\,x}}{\mathrm{e}^{3x} + \mathrm{e}^{-3\,x}} \right) \left(...

And why are you bothering to use a calculator? Surely they would want an exact answer...

Surely it's a trapezium and a rectangle...

I'm asking, why are you even bothering to change the base at all? The base of your exponential equation is 3, surely the best base for your logarithm would therefore also be 3...

Since when is a semicircle a line segment?

Why not just $\displaystyle x = \frac{1}{2}\log_3{\left( 105 \right) }$?

Solve x(x + 4) = 96 for x (which is the width).

That substitution would be fine, but I would probably lean more towards a hyperbolic substitution (just personal preference). $\displaystyle \begin{align*} x = 4\cosh{ \left( t \right) } \implies \mathrm{d}x = 4\sinh{\left( t \right) } \,\mathrm{d}t \end{align*}$

Which is a CAS...