The area between the curve y = 4 + 3x - x2 and x-axis is

Question

The area between the curve y = 4 + 3x - x2 and x-axis is (A) (125/6) sq. units (B) (125/3) sq. units (C) (125/2) sq. units (D) none of these

Tardigrade Admin · Accepted Answer

We have, y = 4 + 3x - x2, a parabola Putting y = 0, we get x2 - 3x - 4 = 0 ⇒ (x-4)(x+1)=0 ⇒ x=-1 or x = 4 ∴ Required area = ∫ limits-14(4+3x-x2)dx =[4x+(3x2/2)-(x3/3)]-14=(125/6) sq. units

Q. The area between the curve $y = 4 + 3 x - x^{2}$ and x-axis is

$\frac{125}{6}$ sq. units

$\frac{125}{3}$ sq. units

$\frac{125}{2}$ sq. units

none of these

We have, $y = 4 + 3 x - x^{2}$ , a parabola
Putting $y = 0$ , we get $x^{2} - 3 x - 4 = 0$
$\Rightarrow (x - 4) (x + 1) = 0 \Rightarrow x = - 1$ or $x = 4$
$∴$ Required area $= - 1 \int 4 (4 + 3 x - x^{2}) d x$
$= [4 x + \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{- 1}^{4} = \frac{125}{6}$ sq. units

Q. The area between the curve y=4+3x−x2 and x-axis is

6125​ sq. units

3125​ sq. units

2125​ sq. units