If A(n) represents the area bounded by the curve y=n log e x(n ∈ N) and n1, x-axis and the lines x =1 and x = e. Then A ( n )- A ( n -1) is equal to

Question

If A(n) represents the area bounded by the curve y=n log e x(n ∈ N) and n>1, x-axis and the lines x =1 and x = e. Then A ( n )- A ( n -1) is equal to (A) 1 (B) n2 (C) ne (D) e

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d1c896aea06495d640e7ec39e73e5635-.png" /> A(n)=∫ limits1e x log e x d x, f(x)=n log e x A(n)=n[x log e x-x]1c =n[e log e-e-(1 log 1-1)]=n text So, A(n)-A(n-1) =n-(n-1)=1

Q. If $A (n)$ represents the area bounded by the curve $y = n lo g_{e} x (n \in N)$ and $n > 1$ , $x$ -axis and the lines $x = 1$ and $x = e$ . Then $A (n) - A (n - 1)$ is equal to

1

$n^{2}$

ne

$e$

$A (n) = 1 \int e x lo g_{e} x d x, f (x) = n lo g_{e} x$
$A (n) = n [x lo g_{e} x - x]_{1}^{c}$
$= n [e lo g e - e - (1 lo g 1 - 1)] = n$
$So, A (n) - A (n - 1)$
$= n - (n - 1) = 1$

Q. If A(n) represents the area bounded by the curve y=nloge​x(n∈N) and n>1, x-axis and the lines x=1 and x=e. Then A(n)−A(n−1) is equal to

1

n2

ne

e

A(n)=1∫e​xloge​xdx,f(x)=nloge​x A(n)=n[xloge​x−x]1c​ =n[eloge−e−(1log1−1)]=n So, A(n)−A(n−1) =n−(n−1)=1

Q. If $A (n)$ represents the area bounded by the curve $y = n lo g_{e} x (n \in N)$ and $n > 1$ , $x$ -axis and the lines $x = 1$ and $x = e$ . Then $A (n) - A (n - 1)$ is equal to

$n^{2}$

$e$

$A (n) = 1 \int e x lo g_{e} x d x, f (x) = n lo g_{e} x$
$A (n) = n [x lo g_{e} x - x]_{1}^{c}$
$= n [e lo g e - e - (1 lo g 1 - 1)] = n$
$So, A (n) - A (n - 1)$
$= n - (n - 1) = 1$