If the area bounded by the curve y = x^3 + ax | vedclass

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

Question

(B) The area $A$ is given by $\int_{-2}^{1} |x^3 + ax| \, dx$. Since $y = x^3 + ax = x(x^2 + a)$, and $a > 0$, the curve crosses the $x$-axis only at

Vedclass · Accepted Answer

$a = 2$

Vedclass · Answer

(B) The area $A$ is given by $\int_{-2}^{1} |x^3 + ax| \, dx$. Since $y = x^3 + ax = x(x^2 + a)$, and $a > 0$, the curve crosses the $x$-axis only at $x = 0$ in the interval $[-2, 1]$. For $x \in [-2, 0]$, $x^3 + ax \le 0$, so $|x^3 + ax| = -(x^3 + ax)$. For $x \in [0, 1]$, $x^3 + ax \ge 0$, so $|x^3 + ax| = x^3 + ax$. Area $= \int_{-2}^{0} -(x^3 + ax) \, dx + \int_{0}^{1} (x^3 + ax) \, dx = \frac{37}{4}$. Evaluating the integrals: $-[\frac{x^4}{4} + \frac{ax^2}{2}]_{-2}^{0} + [\frac{x^4}{4} + \frac{ax^2}{2}]_{0}^{1} = \frac{37}{4}$. $-(0 - (4 + 2a)) + (\frac{1}{4} + \frac{a}{2}) = \frac{37}{4}$. $4 + 2a + \frac{1}{4} + \frac{a}{2} = \frac{37}{4}$. $\frac{5a}{2} = \frac{37}{4} - \frac{17}{4} = \frac{20}{4} = 5$. $a = 2$.

If the area bounded by the curve $y = x^3 + ax$ (where $a > 0$), the $x$-axis, and the lines $x = -2$ and $x = 1$ is $\frac{37}{4}$ square units, find the value of $a$.

Similar Questions

Area of the region bounded by the ellipse $9x^2 + 16y^2 = 1$ in the first quadrant is . . . . . . sq. units.

For which of the following values of $m$,the area of the region bounded by the curve $y = x - x^2$ and the line $y = mx$ equals $\frac{9}{2}$?

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

The area of the region bounded by $y=x^3$, the $x$-axis, $x=-2$, and $x=4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module