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This integral is a particularly difficult integral that has shown up in STEP questions before. It's not hard because of how difficult the actual calculations are, but more because of how hard it is to spot the 'trick' to doing this integral.

The trick is integration by parts twice. You can integrate by parts by letting u=e^x and v'=sin(x). You end up with the integral of e^x cos(x), which you integrate with exactly the same method. Then, you end up with, if I is the integral of e^x sin(x),

I=e^x sin(x) - e^x cos(x) - I + c, which implies that I=e^x (sin(x)-cos(x)) + c.

Answered by Lawrence H.

Studies Mathematics at Cambridge

Suppose p and q are rational. Then

p = p_{1}/p_{2} , q = q_{1}/q_{2 }, where p_{1}, p_{2}, q_{1}, q_{2} are integers.

So pq = p_{1}q_{1}/p_{2}q_{2} , p + q = (p_{1}q_{2} + p_{2}q_{1})/p_{2}q_{2 }, are clearly both rational.

Hence if pq, or p + q, where irrational, then it cannot have been the case that p and q were both rational. Hence 1, or both of p and q, are irrational.

Answered by Ross B.

Studies Mathematics at Bristol

Let f be a function defined on the positive real axis with values in R, such that f(x) = x - ln(1+x). Differentiating this function, we obtain, f'(x) = (x - ln(1+x))' = x' - ln'(1+x) = 1 - 1/(1+x). Since, x > 0, we have 1 + x > 1, and so 1/(1+x) < 1. So, 1 - 1/(1+ x) > 0. So, f'(x) > 0, for all positive x. So, by L'Hospital Rule, f(x) is strictly increasing. Thus, f(x) > lim f(x) when x -> 0 = 0 - ln (1+0) = 0. So, f(x) > 0, for all positive. x.

Answered by Andreea I.

Studies Mathematics and Computer Science at Oxford, Merton College

Well, as with all STEP questions, each problem is unique. But with graphing problems there are some nice things you can do to help break down the problem. Firstly, ALWAYS DIFFERENTIATE, i cannot stress this enough. If you know how your gradient is changing, this gives a lot about what a graph looks like. Secondly, look for roots to both you equation and its derivative. This is a BIG part of these questions and where the most marks will be awarded. Find where your graph hits the y and x axis, and find where the fixed points are. Finally, once you have all your fixed points, make sure to note if they are a maximum, minimum or a turning point.

Now, you should have a pretty good idea of how your graph looks, but there's still one more thing we can do. Find what your graph looks like as it tends to +/- infinity

For example as y = e^x - x^3 tends to infinity, e^x will increase much faster than x^3, so will look a lot like e^x for large x

But for large negative x, e^x is essentially 0, so the graph will look pretty much like x^3

Answered by Oliver M.

Studies MMaths at Warwick

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Version: | 3.55.0 |

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