Let A denote the area bounded by the curve y= | vedclass

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

Question

(D) The area $A$ is given by the integral $\int_{1}^{10} \frac{1}{x} dx = [\ln x]_{1}^{10} = \ln 10$.Consider the sum $B = \sum_{n=1}^{9} \frac{1}{n}

Vedclass · Accepted Answer

$C < A < B$ and $B - A < A - C$

Vedclass · Answer

(D) The area $A$ is given by the integral $\int_{1}^{10} \frac{1}{x} dx = [\ln x]_{1}^{10} = \ln 10$.Consider the sum $B = \sum_{n=1}^{9} \frac{1}{n} = 1 + \frac{1}{2} + \dots + \frac{1}{9}$. This represents the upper Riemann sum for the function $f(x) = \frac{1}{x}$ on the interval $[1, 10]$ with subintervals of width $1$. Since $f(x)$ is a decreasing function,the upper sum is greater than the area under the curve,so $B > A$.Consider the sum $C = \sum_{n=2}^{10} \frac{1}{n} = \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{10}$. This represents the lower Riemann sum for the function $f(x) = \frac{1}{x}$ on the interval $[1, 10]$ with subintervals of width $1$. Since $f(x)$ is decreasing,the lower sum is less than the area under the curve,so $C Thus,we have $C To compare $B-A$ and $A-C$,note that $B-A = \sum_{n=1}^{9} \frac{1}{n} - \int_{1}^{10} \frac{1}{x} dx = \sum_{n=1}^{9} (\frac{1}{n} - \int_{n}^{n+1} \frac{1}{x} dx)$.Similarly,$A-C = \int_{1}^{10} \frac{1}{x} dx - \sum_{n=2}^{10} \frac{1}{n} = \sum_{n=1}^{9} (\int_{n}^{n+1} \frac{1}{x} dx - \frac{1}{n+1})$.Since the function $f(x) = \frac{1}{x}$ is convex,the area under the curve is closer to the lower sum than the upper sum,implying $A-C Therefore,$C < A < B$ and $B - A < A - C$.

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,

Similar Questions

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

The area of the region lying in the first quadrant bounded by $y=4x^2$,$x=0$,$y=2$,and $y=4$ is

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is $24$ sq. units. Then,$a = $ . . . . . . .

$AOB$ is the positive quadrant of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ in which $OA=5, OB=3$. The area between the arc $AB$ and the chord $AB$ of the ellipse in sq. units is

Vedclass Test Series

Exam Paper Generator

Online Exam Module