Q.Find the value of the following: Area bounded by the curve $y = x^3$, the $x$-axis and the ordinates $x = -2$ and $x = 1$ is (A) $-9$ (B) $-\frac{15}{4}$ (C) $\frac{15}{4}$ (D) $\frac{17}{4}$

Concept: Area under a curve — but since $y = x^3$ is negative for $x < 0$, area must be taken as the integral of $|y|$, not the signed integral.

Step 1: The curve crosses the $x$-axis at $x = 0$. For $x \in [-2, 0]$, $x^3 \le 0$; for $x \in [0, 1]$, $x^3 \ge 0$. So total area is:

Step 2: Compute each part:

$\int_{-2}^{0} -x^3 \, dx = -\left[ \frac{x^4}{4} \right]_{-2}^{0} = -\left(0 - \frac{16}{4}\right) = 4$ …

When you’re asked for the area bounded by a curve and the x‑axis, the answer must be positive — area is a geometric quantity, not a signed accumulation. The trap here is that the curve $y = x^3$ changes sign at $x = 0$: it’s negative for $x < 0$ and positive for $x > 0$. If you simply integrate $x^3$ from $-2$ to $1$, you’ll get a net signed area that cancels part of the negative region with the positive region, giving a misleading result.

The correct approach is to split the interval at the point where the curve crosses the axis, take the absolute value of each piece, and add them.

1. Find where the curve meets the x‑axis.

   $y = x^3 = 0$ gives $x = 0$. So the curve is below the axis on $[-2, 0]$ and above it on $[0, 1]$.

2. Set up the area as a sum of absolute integrals.

On $[-2, 0]$, $x^3 \le 0$, so $|x^3| = -x^3$.

On $[0, 1]$, $x^3 \ge 0$, so $|x^3| = x^3$.

3. Compute the first piece.

The antiderivative of $x^3$ is $\frac{x^4}{4}$, so

4. Compute the second piece. …

Method: Area of a curve that changes sign across the x-axis

Use this for the area bounded by $y = f(x)$ and the $x$-axis when $f$ is negative on part of the interval and positive on another (e.g. $y = x^3$ crossing at the origin).

Steps

Step 1: Locate the crossing(s).

Solve $f(x) = 0$ within $[a,b]$. For $y = x^3$ the only crossing is $x = 0$.

Step 2: Split at each crossing and take absolute values.

On the interval where $f \le 0$, replace $|f|$ by $-f$; where $f \ge 0$, keep $f$. …

Common Mistakes

Mistake 1: Integrating $x^3$ straight from $-2$ to $1$.

Why it's wrong: this gives the net signed area $[\tfrac{x^4}{4}]_{-2}^1 = \tfrac14 - 4 = -\tfrac{15}{4}$ — precisely the trap options (A) and (B). Area cannot be negative. Correct approach: split at the crossing $x=0$ and add magnitudes.

Mistake 2: Forgetting the curve is below the axis on $[-2,0]$. …