Notes: Use the properties of integrals to verify the inequality without evaluating the integrals. integrate from `int_0^1 sqrt(1+x^2)dx lt int

Thursday, 15 May 2014

Use the properties of integrals to verify the inequality without evaluating the integrals. integrate from $\displaystyle{\int_{{0}}^{{1}}}\sqrt{{{1}+{{x}}^{{2}}}}{\left.{d}{x}\right.}\lt{\int_{{0}}^{{1}}}\ldots$

You should
remember that $\displaystyle{a}$ dx.

Notice that 0<1 and you need to verify if $\displaystyle\sqrt{{{1}+{{x}}^{{2}}}}<\sqrt{{{1}+{x}}}$
$\displaystyle{o}{v}{e}{r}{t}{h}{e}\int{e}{r}{v}{a}{l}{\left[{0},{1}\right]}{s}{u}{c}{h}{t}{\hat{:}}$

If $\displaystyle{x}\in{\left[{0},{1}\right]}\Rightarrow{{x}}^{{2}}<{x}\Rightarrow$
x^2 + 1 < x + 1 => sqrt(x^2 + 1)

Since $\displaystyle\sqrt{{{{x}}^{{2}}+}}$
1)

$\displaystyle\sqrt{{{{x}}^{{2}}+{1}}}$ int_0^1 sqrt(x^2 + 1) dx < int_0^1
sqrt(x+1)dx

Hence, using the properties of integrals yields
that inequality $\displaystyle{\int_{{0}}^{{1}}}\sqrt{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}<{\int_{{0}}^{{1}}}\sqrt{{{x}+{1}}}{\left.{d}{x}\right.}$
holds.

Notes

Thursday, 15 May 2014

Use the properties of integrals to verify the inequality without evaluating the integrals. integrate from $\displaystyle{\int_{{0}}^{{1}}}\sqrt{{{1}+{{x}}^{{2}}}}{\left.{d}{x}\right.}\lt{\int_{{0}}^{{1}}}\ldots$

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