The area enclosed by the curve y = (x - 1)(x | vedclass

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

Question

(C) The curve is $y = (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$.The area is bounded by the $y$-axis $(x=0)$,the $x$-axis $(y=0)$,and the line $x=3

Vedclass · Accepted Answer

$\frac{11}{4}$

Vedclass · Answer

(C) The curve is $y = (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$.The area is bounded by the $y$-axis $(x=0)$,the $x$-axis $(y=0)$,and the line $x=3$.The roots of the curve are $x=1, 2, 3$. The curve intersects the $x$-axis at these points.Area $A = \int_{0}^{3} |y| dx = \int_{0}^{1} |x^3 - 6x^2 + 11x - 6| dx + \int_{1}^{2} |x^3 - 6x^2 + 11x - 6| dx + \int_{2}^{3} |x^3 - 6x^2 + 11x - 6| dx$.Let $I(x) = \int (x^3 - 6x^2 + 11x - 6) dx = \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x$.$1$. For $x \in [0, 1]$,$y $2$. For $x \in [1, 2]$,$y > 0$,so Area $A_2 = I(2) - I(1) = [(\frac{16}{4} - 2(8) + \frac{11(4)}{2} - 6(2)) - (-\frac{9}{4})] = [(4 - 16 + 22 - 12) + \frac{9}{4}] = [-2 + \frac{9}{4}] = \frac{1}{4}$.$3$. For $x \in [2, 3]$,$y Total Area $= A_1 + A_2 + A_3 = \frac{9}{4} + \frac{1}{4} + \frac{1}{4} = \frac{11}{4}$.

The area enclosed by the curve $y = (x - 1)(x - 2)(x - 3)$ between the coordinate axes and the ordinate at $x = 3$ is:

Similar Questions

If $A$ is the area of the region bounded by the curve $y = \sqrt{3x + 4}$,the $x$-axis,and the lines $x = -1$ and $x = 4$,and $B$ is the area bounded by the curve $y^2 = 3x + 4$,the $x$-axis,and the lines $x = -1$ and $x = 4$,then $A:B$ is equal to:

The area of the region bounded by the curve $y=x^2$ and the line $y=16$ is

Area of the region bounded by the curve $y^{2}=4x$,$y$-axis and the line $y=3$ is

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

The area bounded by the curve $x = 2 - y - y^2$ and the $Y$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module