The area between the curve y = 4 + 3x - x^2 a | vedclass

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

Question

(A) To find the area between the curve $y = 4 + 3x - x^2$ and the $x$-axis,we first find the points of intersection by setting $y = 0$.$4 + 3x - x^2 =

Vedclass · Accepted Answer

$125/6$

Vedclass · Answer

(A) To find the area between the curve $y = 4 + 3x - x^2$ and the $x$-axis,we first find the points of intersection by setting $y = 0$.$4 + 3x - x^2 = 0$$x^2 - 3x - 4 = 0$$(x - 4)(x + 1) = 0$Thus,the intersection points are $x = -1$ and $x = 4$.Since the curve lies above the $x$-axis in the interval $[-1, 4]$,the area is given by the integral:$	ext{Area} = \int_{-1}^{4} (4 + 3x - x^2) dx$$= [4x + \frac{3x^2}{2} - \frac{x^3}{3}]_{-1}^{4}$$= (4(4) + \frac{3(16)}{2} - \frac{64}{3}) - (4(-1) + \frac{3(1)}{2} - \frac{-1}{3})$$= (16 + 24 - \frac{64}{3}) - (-4 + \frac{3}{2} + \frac{1}{3})$$= (40 - \frac{64}{3}) - (-4 + \frac{11}{6})$$= \frac{120 - 64}{3} - (\frac{-24 + 11}{6})$$= \frac{56}{3} - (\frac{-13}{6})$$= \frac{112 + 13}{6} = \frac{125}{6}$

The area between the curve $y = 4 + 3x - x^2$ and the $x$-axis is:

Similar Questions

The volume of the solid generated by revolving the region bounded by the parabolas $y=x^{2}$ and $x=y^{2}$ about the $y$-axis is:

Area of the region bounded by $x^2 = 4y$,the $X$-axis,and the line $x = 3$ is . . . . . . sq. units.

The area of the region (in sq. units) bounded by the curve $y = 2\sqrt{1 - x^2}$ and the $X$-axis is . . . . . . .

Find the area of the region bounded by the curve $y^{2}=x$,the lines $x=1$,$x=4$,and the $x$-axis in the first quadrant.

Area enclosed by the parabola $ay = 3(a^2 - x^2)$ and $x$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module