Find the definite integral of int_{a}^{b} fra | vedclass

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

Question

(A) The integral of $\frac{1}{x}$ with respect to $x$ is $\log_{e} |x|$.Applying the fundamental theorem of calculus for the definite integral $\int_{

Vedclass · Accepted Answer

$\log_{e} \left( \frac{b}{a} \right)$

Vedclass · Answer

(A) The integral of $\frac{1}{x}$ with respect to $x$ is $\log_{e} |x|$.Applying the fundamental theorem of calculus for the definite integral $\int_{a}^{b} \frac{1}{x} dx$:$= [\log_{e} |x|]_{a}^{b}$$= \log_{e} |b| - \log_{e} |a|$Using the logarithmic property $\log_{e} m - \log_{e} n = \log_{e} \left( \frac{m}{n} \right)$:$= \log_{e} \left( \frac{b}{a} \right)$.

Find the definite integral of $\int_{a}^{b} \frac{1}{x} dx$.

Similar Questions

The value of the integral $\int_0^2 \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x}+1)(\sqrt{x^4+x^2+1})} dx$ is equal to:

If $I = \int_{0}^{\frac{\pi}{6}} \frac{\cos x}{x} dx$ and $J = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos x}{x} dx$,which of the following is $CORRECT$?

$\int_{-4}^5 \frac{1}{\sqrt{20+x-x^2}} dx=$

$[x]$ represents the greatest integer function. If $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx = a + b\sqrt{2} + c\sqrt{3}$,then $a + b + c =$

$\int_0^2 |2x - 3| \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module