Q.The area of the region bounded by the curve $y=x^2$, $X$-axis, $x=-1$ and $x=1$ is ____ square unit(s). Choices given: $[\frac43,\ \frac23,\ \frac12,\ 1]$

The key here is to understand what $y = x|x|$ actually looks like. The absolute value on $x$ splits the definition into two cases:

• When $x \ge 0$, $|x| = x$, so $y = x \cdot x = x^2$.
• When $x < 0$, $|x| = -x$, so $y = x \cdot (-x) = -x^2$.

So the curve is a parabola opening upward on the right side, and a parabola opening downward on the left side. It is an odd function: $f(-x) = -f(x)$. This symmetry is the heart of the problem.

The question asks for the area bounded by the curve, the $x$-axis, and the vertical lines $x = -1$ and $x = 1$. "Area bounded" means geometric area — always positive — not signed area (the integral). This is a classic trap.

Let’s work through it.

1. Set up the absolute area integral. The geometric area between a curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$ is $\int_a^b |f(x)| \, dx$. Here:

2. Simplify $|x|x||$. Since $|x|x|| = |x| \cdot |x| = |x|^2 = x^2$ (because squaring removes the sign), we have:

That’s a neat simplification: the absolute value of the function is just $x^2$, a simple upward parabola.

3. Compute the integral.

The antiderivative of $x^2$ is $\frac{x^3}{3}$. So: …

The parabola $y = x^2$ opens upward with vertex at the origin, and the horizontal line $y = 16$ cuts it at two points. Finding those intersection points: $x^2 = 16$ gives $x = \pm 4$. So the bounded region sits between $x = -4$ and $x = 4$, below the line and above the parabola.

The key decision is whether to integrate with respect to $x$ (vertical slices) or $y$ (horizontal slices). Both are valid, but the setup differs.

Vertical slices (integrating with respect to $x$):

At any $x$ between $-4$ and $4$, a vertical strip runs from the parabola $y = x^2$ up to the line $y = 16$. The height of that strip is $16 - x^2$. The area is

Because the integrand $16 - x^2$ is even (symmetric about $x = 0$), this equals

Notice that $\int_{0}^{4} x^2\, dx$ alone is not the area; it gives the area under the parabola from $0$ to $4$, not the region between the parabola and the line. So option (A) is incorrect, and option (B) is also incorrect (it's twice the area under the parabola, not the region we want).

Horizontal slices (integrating with respect to $y$):

At any height $y$ between $0$ and $16$, a horizontal strip extends from the left branch of the parabola to the right branch. Solving $y = x^2$ for $x$ gives $x = \pm \sqrt{y}$. The width of the strip is

The area is then

This matches option (D). …

The problem asks for the area of the region bounded by $y^2 = x$, the vertical line $x = 4$, and the $x$-axis. Let’s first picture what’s happening.

The curve $y^2 = x$ is a right-opening parabola with its vertex at the origin. For a given $x$, $y = \pm \sqrt{x}$. The $x$-axis is $y=0$, and the line $x=4$ cuts off the region on the right. The “shaded region” is typically the part above the $x$-axis — that is, the area under the upper half of the parabola from $x=0$ to $x=4$.

So we want the area between $y = \sqrt{x}$ (the upper branch), the $x$-axis, and the vertical line $x=4$.

1. Set up the integral with respect to $x$

The upper boundary is $y = \sqrt{x}$, the lower boundary is $y = 0$, and $x$ runs from $0$ to $4$. The area is:

That’s exactly option (D).

2. Check the other options

• (A) $\int_{0}^{4} x \, dx$ would give the area under the line $y = x$, not under $\sqrt{x}$. That’s a different shape entirely.
• (B) $\int_{0}^{2} y^2 \, dy$ comes from rewriting $x = y^2$ and integrating with respect to $y$ from $y=0$ to $y=2$ (since at $x=4$, $y=2$). That actually gives the same numerical area — but the question asks for the area of the region bounded by the given curves and the $x$-axis, and the standard representation in $x$ is $\int \sqrt{x} \, dx$. Option (B) is a valid alternative form, but it’s not the one listed that matches the direct $x$-integral.
• (C) $2 \int_{0}^{4} \sqrt{x} \, dx$ would give the area of the full parabola (both upper and lower halves), which is twice the shaded region. …

The question asks for the area of a region bounded by the parabola $y = x^2$ (from $x = 0$ to $x = 2$) and the $y$-axis. The key is recognizing that we can compute area by integrating either horizontally or vertically, and the setup depends entirely on which variable we choose as our integration variable.

When we integrate with respect to $x$, we sum vertical strips of width $dx$ and height $y = x^2$. When we integrate with respect to $y$, we sum horizontal strips of width $dy$ and length equal to the horizontal distance from the $y$-axis to the curve.

Let me identify what happens at the boundaries. At $x = 0$, we have $y = 0^2 = 0$. At $x = 2$, we have $y = 2^2 = 4$. So as $x$ ranges from $0$ to $2$, the variable $y$ ranges from $0$ to $4$.

Now let's examine each option:

1. Option (A): $\displaystyle\int_0^2 x^2\,dx$

   This integrates with respect to $x$ from $0$ to $2$, summing vertical strips of height $x^2$. This gives the area under the curve $y = x^2$ from $x = 0$ to $x = 2$, which is indeed the region described. This is a valid representation.

2. Option (B): $\displaystyle\int_0^2 \sqrt{y}\,dy$

   This integrates with respect to $y$, but only from $0$ to $2$. Since the curve reaches $y = 4$ when $x = 2$, stopping at $y = 2$ would only capture part of the region. This is incorrect.

3. Option (C): $\displaystyle\int_0^4 x^2\,dx$

   This integrates with respect to $x$ from $0$ to $4$, which extends beyond the given domain $0 \le x \le 2$. This would compute the area under the parabola all the way to $x = 4$, which is not our region. This is incorrect.

4. Option (D): $\displaystyle\int_0^4 \sqrt{y}\,dy$ …