STEP II 1997 question 8 solution

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STEP II 1997 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1997 question 8 solution

I'm not sure what the "standard explanation" is here. I think Siklos says you just need to draw a couple of vague wiggles and say the integral is the area under the curve. More formally, if f,g are cts, consider the integral of f-g, apply the fundamental theorem of calculus and the mean value theorem to show that $\displaystyle \int_a^b (f-g)(x) dx = (b-a)[f(c)-g(c)]$ for some $\displaystyle c \in (a,b)$ and so is greater than 0.

If p > q and x>=1 then $\displaystyle x^p \ge x^q$ . So $\displaystyle \int_1^x t^p dt \ge \int_1^x t^q dt$ . So $\displaystyle \frac{x^p-1}{p} \ge \frac{x^q-1}{q}$ as desired.

If p > q and 0 <= x <1 then $\displaystyle x^p \le x^q$ . So $\displaystyle \int_x^1 t^p dt \le \int_x^1 t^q dt$ and so $\displaystyle \frac{1-x^p}{p} \le \frac{1-x^q}{q}$ . Multiplying by -1 confirms the earlier inequality for x >=0.

Doing the same trick on the inequality we've just found gives us $\displaystyle \int_0^x \frac{1}{p} (x^p - 1) dx \ge \int_0^x \frac{1}{q} (x^q - 1) dx$ and so