If f(alpha) = int_{1}^{alpha} frac{log_{10} t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(\alpha) = \int_{1}^{\alpha} \frac{\ln t}{(\ln 10)(1+t)} dt$.$f(e^{3}) = \int_{1}^{e^{3}} \frac{\ln t}{(\ln 10)(1+t)} dt \quad \dots(1)$Fo

Vedclass · Accepted Answer

$\frac{9}{2 \log_{e}(10)}$

Vedclass · Answer

(D) Given $f(\alpha) = \int_{1}^{\alpha} \frac{\ln t}{(\ln 10)(1+t)} dt$.$f(e^{3}) = \int_{1}^{e^{3}} \frac{\ln t}{(\ln 10)(1+t)} dt \quad \dots(1)$For $f(e^{-3}) = \int_{1}^{e^{-3}} \frac{\ln t}{(\ln 10)(1+t)} dt$,let $t = \frac{1}{x}$,so $dt = -\frac{1}{x^{2}} dx$.When $t=1, x=1$ and when $t=e^{-3}, x=e^{3}$.$f(e^{-3}) = \int_{1}^{e^{3}} \frac{\ln(1/x)}{(\ln 10)(1+1/x)} \left(-\frac{1}{x^{2}}\right) dx = \int_{1}^{e^{3}} \frac{-\ln x}{(\ln 10)(\frac{x+1}{x})} \left(-\frac{1}{x^{2}}\right) dx = \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln x}{x(x+1)} dx \quad \dots(2)$Adding $(1)$ and $(2)$:$f(e^{3}) + f(e^{-3}) = \frac{1}{\ln 10} \int_{1}^{e^{3}} \left( \frac{\ln t}{1+t} + \frac{\ln t}{t(1+t)} \right) dt$$= \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{1+t} \left( 1 + \frac{1}{t} \right) dt = \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{1+t} \left( \frac{t+1}{t} \right) dt$$= \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{t} dt$Let $\ln t = u$,then $\frac{1}{t} dt = du$. When $t=1, u=0$ and when $t=e^{3}, u=3$.$= \frac{1}{\ln 10} \int_{0}^{3} u du = \frac{1}{\ln 10} \left[ \frac{u^{2}}{2} \right]_{0}^{3} = \frac{1}{\ln 10} \left( \frac{9}{2} \right) = \frac{9}{2 \log_{e} 10}$.

If $f(\alpha) = \int_{1}^{\alpha} \frac{\log_{10} t}{1+t} dt, \alpha > 0$,then $f(e^{3}) + f(e^{-3})$ is equal to.

Similar Questions

If $I_{n} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \, dx$,where $n$ is a positive integer,then $I_{10} + I_{8}$ is equal to

$\int_{-2}^{\pi} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \,dx$ is equal to (where $[\cdot]$ denotes the greatest integer function).

The value of $\int_0^1 \tan^{-1}(1-x+x^2) dx$ is

Let $f$ be a positive function. Let $I_1 = \int_{1 - k}^k x f\{x(1 - x)\} dx$ and $I_2 = \int_{1 - k}^k f\{x(1 - x)\} dx$,where $2k - 1 > 0$. Then $I_1/I_2$ is

Let $f:R \to R$ and $g:R \to R$ be continuous functions,then the value of the integral $\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module