Question

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

• A. 64
• B. 81/4
• C. 66/5
• D. 68

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).

Answer 1

The Correct Option is D

Step 1: Set up the definite integral for the area.
The area of the region bounded by a curve $y=f(x)$, the x-axis, and the lines $x=a$ and $x=b$ is given by $A = \int_a^b |f(x)| dx$. Here, $f(x)=x^3$, $a=-2$, and $b=4$. \[ A = \int_{-2}^{4} |x^3| dx. \] 

Step 2: Split the integral based on the sign of the function. 
The function $f(x)=x^3$ is negative for $x<0$ and positive for $x>0$. We need to split the integral at $x=0$. \[ |x^3| = \begin{cases} -x^3, & \text{if } x<0 
x^3, & \text{if } x \ge 0 \end{cases}. \] The integral becomes: \[ A = \int_{-2}^{0} (-x^3) dx + \int_{0}^{4} (x^3) dx. \] 

Step 3: Evaluate the first integral. 
\[ \int_{-2}^{0} (-x^3) dx = \left[ -\frac{x^4}{4} \right]_{-2}^{0}. \] \[ = \left( -\frac{0^4}{4} \right) - \left( -\frac{(-2)^4}{4} \right) = 0 - \left( -\frac{16}{4} \right) = 4. \] 

Step 4: Evaluate the second integral. 
\[ \int_{0}^{4} (x^3) dx = \left[ \frac{x^4}{4} \right]_{0}^{4}. \] \[ = \left( \frac{4^4}{4} \right) - \left( \frac{0^4}{4} \right) = \frac{256}{4} - 0 = 64. \] 

Step 5: Calculate the total area. 
The total area is the sum of the areas from the two parts. \[ A = 4 + 64 = 68. \] \[ \boxed{A=68}. \]