f(α)=\(\int_{1}^{\alpha}\)\(\frac{log 10t}{1+t}\) dt⋯(i) f(\(\frac{1}{\alpha}\))=\(\int_{1}^{\frac{1}{\alpha}}\frac{log_{10}t}{1+t}dt\) Substituting t = \(\frac{1}{p}\) f(\(\frac{1}{\alpha}\))=\(\int_{1}^{\frac{1}{\alpha}}\frac{log_{10}(\frac{1}{p})}dt\) =\(\int_{1}^{\alpha} \frac{log_{10}p}{p(p+1)}dp\)=\(\int_{1}^{\alpha}(log \frac{10t }{t}-\frac{log\,10t}{t}+1)\)dt ⋯(ii) By (i) + (ii) f(α)+f(\(\frac{1}{\alpha}\))=\(\int_{1}^{\alpha}\)\(\frac{log\,10t}{t}\) dt=\(\int_{1}^{\alpha}\)\(\frac{ln\,t}{t.log_{10}e}\)dt α=e3⇒f(e3)+f(e−3)=\(\frac{9}{2}log_e10\) So, the correct option is (D): \(\frac{9}{2}log_e10\)
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
