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

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

Question

(C) The area $A$ bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$.Given $y =

Vedclass · Accepted Answer

$(\log 4 - 1) 	ext{ sq. units}$

Vedclass · Answer

(C) The area $A$ bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$.Given $y = \log x$,$x = 1$,and $x = 2$.Since $\log x > 0$ for $x \in [1, 2]$,the area is $\int_{1}^{2} \log x \, dx$.Using integration by parts,$\int \log x \, dx = x \log x - x + C$.Thus,$A = [x \log x - x]_{1}^{2}$.$A = (2 \log 2 - 2) - (1 \log 1 - 1)$.Since $\log 1 = 0$,$A = 2 \log 2 - 2 + 1 = 2 \log 2 - 1$.Using the property $n \log m = \log(m^n)$,$2 \log 2 = \log(2^2) = \log 4$.Therefore,$A = (\log 4 - 1) 	ext{ sq. units}$.

The area bounded by the curve $y = \log x$,the $x$-axis,and the ordinates $x = 1$ and $x = 2$ is:

Similar Questions

Let $A$ denote the area bounded by the curve $y=\frac{1}{x}$ and the lines $y=0, x=1, x=10$. Let $B=1+\frac{1}{2}+\ldots+\frac{1}{9}$ and $C=\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{10}$. Then,

Area of the region bounded by $x^2 = 4y$,the $X$-axis,and the line $x = 3$ is . . . . . . sq. units.

The area bounded by the curve $x^2 = 8y$ and the straight line $x - 8y + 2 = 0$ is

The area (in sq. units) bounded by the curves $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module