Area included between curves y = x2 -3 x + 2 and y = - x2 + 3x - 2 is

Question

Area included between curves y = x2 -3 x + 2 and y = - x2 + 3x - 2 is (A) (1/6) sq unit (B) (1/2) sq unit (C) 1 sq unit (D) (1/3) sq unit

Tardigrade Admin · Accepted Answer

Required area <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d6d759ec475321357f5d3642aa99aa3e-.jpeg" /> =2∫ limits21 (-x2+3x-2)dx =2[(-x3/3)+(3x2/2)-2x]21 =2[(-8/3)+6-4+(1/3)-(3/2)+2] =2[(-8/3)+4-(7/6)] =2⋅(1/6)=(1/3) sq unit

Q. Area included between curves $y = x^{2} - 3 x + 2$ and $y = - x^{2} + 3 x - 2$ is

$\frac{1}{6}$ sq unit

$\frac{1}{2}$ sq unit

$1$ sq unit

$\frac{1}{3}$ sq unit

Required area

$= 2 1 \int 2 (- x^{2} + 3 x - 2) d x$

$= 2 [\frac{- x ^{3}}{3} + \frac{3 x ^{2}}{2} - 2 x]_{1}^{2}$

$= 2 [\frac{- 8}{3} + 6 - 4 + \frac{1}{3} - \frac{3}{2} + 2]$

$= 2 [\frac{- 8}{3} + 4 - \frac{7}{6}]$

$= 2 \cdot \frac{1}{6} = \frac{1}{3}$ sq unit

Q. Area included between curves y=x2−3x+2 and y=−x2+3x−2 is

61​ sq unit

21​ sq unit

1 sq unit

31​ sq unit