The area bounded by the curve y = log x betwe | vedclass

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

Question

(B) We have the curve $y = \log x$. To find the limits of integration,we find the intersection with the $x$-axis by setting $y = 0$: $0 = \log x \impl

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We have the curve $y = \log x$. To find the limits of integration,we find the intersection with the $x$-axis by setting $y = 0$: $0 = \log x \implies x = 1$. The area $A$ bounded by the curve between $x = 1$ and $x = e$ is given by the integral: $A = \int_{1}^{e} \log x \, dx$. Using integration by parts,$\int \log x \, dx = x \log x - x$. Evaluating the definite integral: $A = [x \log x - x]_{1}^{e} = (e \log e - e) - (1 \log 1 - 1)$. Since $\log e = 1$ and $\log 1 = 0$: $A = (e(1) - e) - (0 - 1) = (e - e) + 1 = 1$. Thus,the area is $1$ square unit.

The area bounded by the curve $y = \log x$ between the $x$-axis and the ordinate $x = e$ is

Similar Questions

Let $A$ be the area bounded by the curve $y=x|x-3|$,the $x$-axis and the ordinates $x=-1$ and $x=2$. Then $12A$ is equal to $...........$.

The area bounded by the parabola ${y^2} = 4ax$ and its latus rectum is:

The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and its latus rectum is . . . . . . sq. units. (where $e$ denotes the eccentricity of the ellipse).

The area of the region bounded by the curve $y = \cos x$,the $X$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

The area bounded by the $x-$ axis,the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $\sqrt{b^2 + 1} - \sqrt{2}$ for all $b > 1$. Then $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module