Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$

Question

Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$

$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$

I tried to prove it but my answer is not correct.

For first part,As $0\leq x\leq2\Rightarrow 2\leq x^2+2\leq6\Rightarrow\frac{1}{6}\leq\frac{1}{x^2+2}\leq\frac{1}{2}\Rightarrow\frac{1}{6}\int_{0}^{2}1dx\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{1}{2}\int_{0}^{2}1dx$
$\Rightarrow\frac{1}{3}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq1$

In the second part,$0\leq x\leq2\Rightarrow -2<x^2-x<4\Rightarrow e^{-2}<e^{x^2-x}<e^4\Rightarrow2e^{-2}<\int_{0}^{2}e^{x^2-x}dx<2e^4$

Where have i gone wrong?What is the correct method to solve it.

Answer 1

$(i)$ Let $$\displaystyle I = \int_{0}^{2}\frac{1}{2+x^2}dx = \int_{0}^{1}\frac{1}{2+x^2}dx+\int_{1}^{2}\frac{1}{2+x^2}dx\geq \int_{1}^{2}\frac{1}{3}dx+\int_{0}^{1}\frac{1}{6}dx \geq \frac{1}{2}$$

And $$\displaystyle I = \int_{0}^{2}\frac{1}{2+x^2}dx = \int_{0}^{1}\frac{1}{2+x^2}dx+\int_{1}^{2}\frac{1}{2+x^2}dx\leq \int_{0}^{1}\frac{1}{2}dx+\int_{1}^{2}\frac{1}{3}dx \leq\frac{5}{6}$$

Answer 2

For the second problem, we note that

$$-\frac14\le x^2-x\le2$$for $x\in[0,2]$. Note that we can easily find the minimum by taking the derivative and setting it to $0$, while we merely check the endpoints to find the maximum.

Inasmuch as the exponential function is monotonically increasing, then

$$e^{-\frac14}\le e^{x^2-x}<e^2 \tag 1$$

Then, we see immediately that

$$2\,e^{-\frac14}\le \int_0^2 e^{x^2-x}\,dx \le 2\,e^2$$

Answer 3

Both answers are fine, I would just point out than since $f(x)=\frac{\sqrt{x}}{2+x}$ is a concave function on $[0,4]$ and $e^{x^2-x}$ is a convex function on $[0,2]$, both inequalities can be improved through the Hermite-Hadamard inequality, i.e. by using the usual rectangle/trapezoid method for approximating an integral.

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About the first integral, another chance is given by the identity $\frac{1}{2+x^2}=\frac{1}{2}-\frac{x}{6}+\frac{x(x-1)(x-2)}{6(2+x^2)}$, from which: $$ I=\int_{0}^{2}\frac{dx}{2+x^2}=\frac{2}{3}+\frac{1}{6}\int_{0}^{2}\frac{x(x-1)(x-2)}{2+x^2}\,dx, $$ and since $x(x-1)(x-2)$ over $[0,2]$ is between $-\frac{2}{3\sqrt{3}}$ and $\frac{2}{3\sqrt{3}}$, $$ I \leq \frac{2}{3}+\frac{1}{9\sqrt{3}}\,I,\qquad I\geq \frac{2}{3}-\frac{1}{9\sqrt{3}}\,I $$ lead to:

$$ \color{red}{\frac{5}{8}}<0.626478\ldots=\frac{2}{3+\frac{1}{3\sqrt{3}}}\leq \color{red}{I} \leq \frac{2}{3-\frac{1}{3\sqrt{3}}}=0.712364\ldots<\color{red}{\frac{5}{7}}. $$

The second integral is clearly concentrated around $x=2$:

$$ J=\int_{0}^{2}e^{x^2-x}\,dx = e^2\int_{0}^{2}e^{-3x+x^2}\,dx\geq e^2\int_{0}^{2}e^{-3x}\,dx= \frac{e^2-e^{-4}}{3},$$ $$ J \leq e^2\int_{0}^{2}e^{-x}\,dx = e^2-1,$$ but we may improve the last inequality up to: $$ J\leq e^{2}\int_{0}^{2}e^{-x}\left(1-\left(1-\frac{1}{e}\right)x(2-x)\right)\,dx = e^2-5+\frac{4}{e} $$ and the first inequality can be improved up to: $$ J\geq e^2\int_{0}^{2}e^{-3x}(1+x^2)\,dx = \frac{11 e^2-59 e^{-4}}{27} $$ hence $J$ is between $2.97$ and $3.87$.