Question:medium

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

Show Hint

When calculating the area bounded by a curve and the x-axis, you must use the absolute value of the function, $|f(x)|$. This often requires splitting the integral into multiple intervals at the points where the function crosses the x-axis (i.e., at its roots).

Updated On: Mar 30, 2026

64
81/4
66/5
68

Show Solution

The Correct Option is D

Solution and Explanation

Was this answer helpful?

0

Top Questions on Integration

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

If \( p_1, p_2, p_3 \) are the altitudes and \( a=4, b=5, c=6 \) are the sides of a triangle ABC, then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

TS EAMCET - 2025
Geometry

If \(\alpha, \beta, \gamma, \delta, \epsilon\) are the roots of the equation \(x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0\) and \(\alpha<\beta<\gamma<\delta<\epsilon\) then \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = \)

TS EAMCET - 2025
Algebra