Let R be the region satisfying yx-1, x0, then area of R is

Question

Let R be the region satisfying y< x2+1, y >x-1, x< 1 and x >0, then area of R is (A) (11/6) (B) (3/2) (C) (5/6) (D) 2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a33f33e396081c88094a21be702160de-.png" /> A =∫ limits01(( x 2+1)-( x -1)) dx =∫ limits01( x 2- x +2) dx =(11/6)

Q. Let $R$ be the region satisfying $y < x^{2} + 1, y > x - 1, x < 1$ and $x > 0$ , then area of $R$ is

$\frac{11}{6}$

$\frac{3}{2}$

$\frac{5}{6}$

2

$A = 0 \int 1 ((x^{2} + 1) - (x - 1)) d x$
$= 0 \int 1 (x^{2} - x + 2) d x$
$= \frac{11}{6}$

Q. Let R be the region satisfying y<x2+1,y>x−1,x<1 and x>0, then area of R is

611​

23​

65​

2

A=0∫1​((x2+1)−(x−1))dx =0∫1​(x2−x+2)dx =611​

Q. Let $R$ be the region satisfying $y < x^{2} + 1, y > x - 1, x < 1$ and $x > 0$ , then area of $R$ is

$\frac{11}{6}$

$\frac{3}{2}$

$\frac{5}{6}$

$A = 0 \int 1 ((x^{2} + 1) - (x - 1)) d x$
$= 0 \int 1 (x^{2} - x + 2) d x$
$= \frac{11}{6}$