OCR C2 Integration

For the first one, split the fraction and use the laws of indices.

For the second one, integrate it like you would normally and then just put a in instead of a numerical value for the upper limit. "a is a constant greater than 2" just means that $a \ge 2$... but that has no effect on this integral -- all it means is that your integral's not screwed up*; notice that if $a<2$ then your upper limit is lower than your lower limit.

*In fact having an upper limit lower than your lower limit is fine, but if this is C2 it's not worth worrying about, which is probably why they imposed the condition a>2.

$\displaystyle \int \frac{x^3+3x^\frac{1}{2}}{x} dx= \displaystyle \int \frac{x^3}{x}+ \frac{3x^\frac{1}{2}}{x}dx=\frac{1}{3}x +$

I'm not so sure on the last term.

You haven't applied the laws of indices; when you integrate $\dfrac{x^3}{x}$ you don't get $\dfrac{x}{3}$, that's for sure. Simplify $\dfrac{x^3}{x}$ and $\dfrac{3x^{\frac{1}{2}}}{x}$ first.

You haven't applied the laws of indices; when you integrate $\dfrac{x^3}{x}$ you don't get $\dfrac{x}{3}$, that's for sure. Simplify $\dfrac{x^3}{x}$ and $\dfrac{3x^{\frac{1}{2}}}{x}$ first.

That's right. So what do you get when you integrate that? [Hint: not x/3]



You can simplify $\dfrac{\sqrt{x}}{x}$. Notice that $x = \sqrt{x} \times \sqrt{x}$. But really you could have just used the laws of indices; you should know from C1 that $\dfrac{x^a}{x^b} = x^{a-b}$.

I'm not doing A level maths. I'm only in year 10 but I find integration interesting

EDIT: I have simplified it further.

Okay, see my edit I think the laws of indices is a GCSE thing anyway, but if you're in Y10 you might not have covered it yet... I don't know what the syllabus is these days.

The answer to the second question is $\frac{a^3}{4} -2$

I can't seem to get that one too.

That's definitely not the right answer to the second bit, assuming you've not made a typo. You should be aiming for $\dfrac{1}{4} - \dfrac{2}{a^3}$.