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.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

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

Similar Questions

The area of the region bounded by the curve $y = x^2 - x$ and the $X$-axis is . . . . . . sq. units.

The values of a function $f(x)$ at different values of $x$ are as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$
$f(x)$ $2$ $3$ $6$ $11$ $18$ $27$

Then, the approximate area (in square units) bounded by the curve $y=f(x)$ and the $x$-axis between $x=0$ and $x=5$, using the Trapezoidal rule, is:

Find the area of the region bounded by the curve $y=x^{2}$ and the line $y=4$. (in $/3$)

The area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$ is:

If $A$ is the area of the region bounded by the curve $y = \sqrt{3x + 4}$,the $x$-axis,and the lines $x = -1$ and $x = 4$,and $B$ is the area bounded by the curve $y^2 = 3x + 4$,the $x$-axis,and the lines $x = -1$ and $x = 4$,then $A:B$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module