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So I have tried to solve this question in two methods:

Given that $g\leqslant\int_{0}^{1} \frac{x}{x^2 +1} dx \leqslant h$, Find g, h.

The first way I tried is defining $f(x) = \frac{x}{x^2+1}$ and then I found its maximum and minimum values using the first derivative method and continued with integration, got that $g=0, h = 0.5$. Then I tried this:

$$0\leqslant x\leqslant 1$$ $$\frac{1}{x}\leqslant \frac{x^2+1}{x}\leqslant \frac{2}{x}$$ $$0.5x\leqslant \frac{x}{x^2+1}\leqslant x$$ $$\int_{0}^{1}0.5x dx\leqslant \int_{0}^{1}\frac{x}{x^2+1} dx\leqslant \int_{0}^{1}x dx $$

and by solving the integrals, $g=0.25, h = 0.5$

Why have I got different values for $g$? What mistakes have I made?

Also, I always had this other question in mind. A definite integral at last is simply a constant and we can know its value, why do we use the maximum and minimum value property of integration if we already know its value? why do we even use "greater(or less) than or equal" sign?

passepartout
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1 Answers1

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You used two distinct approximations and so it is natural that you got two distinct values.

Anyway, since$$\int_0^1\frac x{x^2+1}\,\mathrm dx=\log\left(\sqrt2\right)\approx0.346574,$$both of your answers are correct.

José Carlos Santos
  • 427,504