Jameson Lucero

Answered

2022-07-06

Evaluating $\underset{x\to 1}{lim}\frac{{\int}_{1}^{x}{\mathrm{ln}}^{2}(t)dt}{(x-1{)}^{3}}$

Answer & Explanation

verzaadtwr

Expert

2022-07-07Added 17 answers

In terms of the evaluation of the limit: applying L'Hôpital's rule once forms

$\underset{x\to 1}{lim}\frac{{\int}_{1}^{x}{\mathrm{ln}}^{2}(t)\phantom{\rule{thinmathspace}{0ex}}dt}{(x-1{)}^{3}}=\underset{x\to 1}{lim}\frac{{\mathrm{ln}}^{2}(x)}{3(x-1{)}^{2}},$

where the chain rule implies that the derivative of the denominator is

$\frac{d}{dx}[(x-1{)}^{3}]=3(x-1{)}^{2}.$

From here you can apply L'Hôpital's rule two more times to find

$\underset{x\to 1}{lim}\frac{{\mathrm{ln}}^{2}(x)}{3(x-1{)}^{2}}=\underset{x\to 1}{lim}\frac{\mathrm{ln}(x)}{3x(x-1)}=\underset{x\to 1}{lim}\frac{1}{x(6x-3)}=\frac{1}{3}.$

To answer your second question, the result follows from the fundamental theorem of calculus. Let f be a continuous, real-valued function defined on the closed interval [a,b]. Suppose F is defined for all $x\in [a,b]$ by

$F(x)={\int}_{a}^{g(x)}f(t)\phantom{\rule{thinmathspace}{0ex}}dt.$

Then

${F}^{\prime}(x)=\frac{d}{dx}{\int}_{a}^{g(x)}f(t)\phantom{\rule{thinmathspace}{0ex}}dt=f(g(x)){g}^{\prime}(x).$

Note that we could change the lower integration constant to any other positive real constant $a>0$ (as the fundamental theorem of calculus would still hold) to deduce

${F}^{\prime}(x)=\frac{d}{dx}{\int}_{a}^{x}{\mathrm{ln}}^{2}(t)\phantom{\rule{thinmathspace}{0ex}}dt={\mathrm{ln}}^{2}(x).$

However, if the upper limit of integration was ${x}^{2}$ instead of x, then we would find

${F}^{\prime}(x)=\frac{d}{dx}{\int}_{a}^{{x}^{2}}{\mathrm{ln}}^{2}(t)\phantom{\rule{thinmathspace}{0ex}}dt=2x{\mathrm{ln}}^{2}({x}^{2}).$

$\underset{x\to 1}{lim}\frac{{\int}_{1}^{x}{\mathrm{ln}}^{2}(t)\phantom{\rule{thinmathspace}{0ex}}dt}{(x-1{)}^{3}}=\underset{x\to 1}{lim}\frac{{\mathrm{ln}}^{2}(x)}{3(x-1{)}^{2}},$

where the chain rule implies that the derivative of the denominator is

$\frac{d}{dx}[(x-1{)}^{3}]=3(x-1{)}^{2}.$

From here you can apply L'Hôpital's rule two more times to find

$\underset{x\to 1}{lim}\frac{{\mathrm{ln}}^{2}(x)}{3(x-1{)}^{2}}=\underset{x\to 1}{lim}\frac{\mathrm{ln}(x)}{3x(x-1)}=\underset{x\to 1}{lim}\frac{1}{x(6x-3)}=\frac{1}{3}.$

To answer your second question, the result follows from the fundamental theorem of calculus. Let f be a continuous, real-valued function defined on the closed interval [a,b]. Suppose F is defined for all $x\in [a,b]$ by

$F(x)={\int}_{a}^{g(x)}f(t)\phantom{\rule{thinmathspace}{0ex}}dt.$

Then

${F}^{\prime}(x)=\frac{d}{dx}{\int}_{a}^{g(x)}f(t)\phantom{\rule{thinmathspace}{0ex}}dt=f(g(x)){g}^{\prime}(x).$

Note that we could change the lower integration constant to any other positive real constant $a>0$ (as the fundamental theorem of calculus would still hold) to deduce

${F}^{\prime}(x)=\frac{d}{dx}{\int}_{a}^{x}{\mathrm{ln}}^{2}(t)\phantom{\rule{thinmathspace}{0ex}}dt={\mathrm{ln}}^{2}(x).$

However, if the upper limit of integration was ${x}^{2}$ instead of x, then we would find

${F}^{\prime}(x)=\frac{d}{dx}{\int}_{a}^{{x}^{2}}{\mathrm{ln}}^{2}(t)\phantom{\rule{thinmathspace}{0ex}}dt=2x{\mathrm{ln}}^{2}({x}^{2}).$

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