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Question: How do you use second fundamental theorem of calculus to find the derivative of given \(\int{\left[ ...

How do you use second fundamental theorem of calculus to find the derivative of given ∫[ln⁡2(t)t]dt\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt from [3, x]?

Explanation

Solution

Now we are given with the integral ∫[ln⁡2(t)t]dt\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt . To solve this integral we will first use the method of substitution. To do so we will substitute ln(t) = u. Now we will get a simplified integral which can be solved by using the standard integral ∫xn=xn+1n+1+C\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C . Now we know that according to second fundamental theorem of calculus we have if F′(x)=f(x)F'\left( x \right)=f\left( x \right) then ∫abf(x)=F(b)−F(a)\int\limits_{a}^{b}{f\left( x \right)}=F\left( b \right)-F\left( a \right) . Hence using this we will get the integral ∫[ln⁡2(t)t]dt\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt from [3, x].

Complete step-by-step solution:
Now first let us solve the given integration ∫[ln⁡2(t)t]dt\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt
To solve the integration we will use a method of substitution.
Let us substitute ln(t) = u.
Now differentiating the above equation on both sides we get,
⇒1tdt=du\Rightarrow \dfrac{1}{t}dt=du
Now we will re-substitute the values of ln(t) and 1tdt\dfrac{1}{t}dt in the given integral. Hence, we get
⇒∫[ln⁡2(t)t]dt=∫u2du\Rightarrow \int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt=\int{{{u}^{2}}du}
Now we know the standard integral ∫xn=xn+1n+1+C\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C Hence using this we get,
∫u2=u33+C\int{{{u}^{2}}}=\dfrac{{{u}^{3}}}{3}+C
Now re-substituting the value of u we get,
⇒∫[ln⁡(t)2t]dt=ln⁡(t)33+C..............(1)\Rightarrow \int{\left[ \dfrac{\ln {{\left( t \right)}^{2}}}{t} \right]dt}=\dfrac{\ln {{\left( t \right)}^{3}}}{3}+C..............\left( 1 \right)
Now by second fundamental theorem of calculus we know that if we have F′(x)=f(x)F'\left( x \right)=f\left( x \right) then ∫abf(x)=F(b)−F(a)\int\limits_{a}^{b}{f\left( x \right)}=F\left( b \right)-F\left( a \right) . Hence using this theorem and equation (1) we get,
⇒∫3xln⁡2(t)tdt=ln⁡3(x)3−ln⁡3(3)3 ⇒∫3xln⁡2(t)tdt=ln⁡3(x)−ln⁡333 \begin{aligned} & \Rightarrow \int_{3}^{x}{\dfrac{{{\ln }^{2}}\left( t \right)}{t}}dt=\dfrac{{{\ln }^{3}}\left( x \right)}{3}-\dfrac{{{\ln }^{3}}\left( 3 \right)}{3} \\\ & \Rightarrow \int_{3}^{x}{\dfrac{{{\ln }^{2}}\left( t \right)}{t}}dt=\dfrac{{{\ln }^{3}}\left( x \right)-{{\ln }^{3}}3}{3} \\\ \end{aligned}
Hence by second fundamental theorem of calculus we get ∫3xln⁡2(t)tdt=ln⁡3(x)−ln⁡333\int_{3}^{x}{\dfrac{{{\ln }^{2}}\left( t \right)}{t}}dt=\dfrac{{{\ln }^{3}}\left( x \right)-{{\ln }^{3}}3}{3}

Note: Now keep a note that if we have F′(x)=f(x)F'\left( x \right)=f\left( x \right) then ∫abf(x)=F(b)−F(a)\int\limits_{a}^{b}{f\left( x \right)}=F\left( b \right)-F\left( a \right) and not ∫abF(x)=f(b)−f(a)\int_{a}^{b}{F\left( x \right)}=f\left( b \right)-f\left( a \right) . Also while using the method of substitution always remember to re-substitute the expression by the original variable. Also while changing variables do not forget to change the derivative of the respective variable. For example if we substitute ln⁡(t)=u\ln \left( t \right)=u then du=1tdtdu=\dfrac{1}{t}dt Hence do not just replace the differential dt by du.