Q.Find the area under the given curves and given lines:

Step 1 — The concept (the trap in this question).

Area is a positive quantity. Where the curve lies below the x-axis, the definite integral is negative, so we must take its modulus:

Blindly writing $\int_{-1}^{2} x\,dx$ lets the negative part cancel part of the positive part — that gives $3/2$, which is why option (A) is there as a distractor.

Step 2 — Find where the sign changes.

$y = x = 0$ at $x = 0$, which lies inside $[-1, 2]$. So split there:

• On $[-1, 0]$: $x < 0$, the line is below the x-axis.
• On $[0, 2]$: $x > 0$, the line is above the x-axis.

Step 3 — Area below the axis, $[-1,0]$.

Step 4 — Area above the axis, $[0,2]$.

Step 5 — Total area.

Geometric check. The region is two right-angled triangles: one with base 1 and height 1 (area $\tfrac12 \cdot 1 \cdot 1 = \tfrac12$) and one with base 2 and height 2 (area $\tfrac12 \cdot 2\cdot 2 = 2$). Total $= \tfrac52$ ✓.

ANSWER: B

Concept & Intuition

We want the area trapped between the downward-opening parabola $y = -x^2$ and the line $x + y + 2 = 0$ (which can be rewritten as $y = -x - 2$). The region is bounded where the line lies above the parabola (since the parabola is more negative for most $x$). To find the area, we:

• Find where they intersect (solve $-x^2 = -x - 2$).
• Determine which curve is on top in the interval between intersections.
• Integrate (top minus bottom) with respect to $x$.

Step-by-step solution

1. Rewrite the line

   $x + y + 2 = 0 \implies y = -x - 2$.

2. Find intersection points

   Set $-x^2 = -x - 2$:

Factor: $(x - 2)(x + 1) = 0$ → $x = -1$ and $x = 2$.

3. Determine which function is greater on $[-1, 2]$

   Test a point, say $x = 0$:

   • Parabola: $y = -0^2 = 0$
   • Line: $y = -0 - 2 = -2$ Since $0 > -2$, the parabola is above the line at $x=0$. But wait — that would mean the parabola is on top? Let’s check the shape: The parabola opens downward, so at $x=0$ it peaks at 0; the line is sloping downward. However, the enclosed region is the one where the line is above the parabola? Let’s test another point: at $x = -1$ (intersection), both equal. At $x = 1$: parabola $y = -1$, line $y = -3$. So parabola is above the line. That means the region between them is bounded above by the parabola and below by the line. But the problem says “area enclosed by the curve and the line” — that region is indeed the one where the parabola is above the line between the intersections. So top = parabola, bottom = line.

4. Set up the integral

   Area = $\int_{x=-1}^{2} [(-x^2) - (-x - 2)] \, dx$

   Simplify the integrand:

5. Integrate

Evaluate at $x=2$:

Evaluate at $x=-1$:

Common denominator 6: $\frac{2}{6} + \frac{3}{6} - \frac{12}{6} = -\frac{7}{6}$

Subtract:

6. Conclusion The area is $4.5$ square units.

The key idea here is that area is always positive. When a curve goes below the x-axis, the definite integral gives a signed area (negative below the axis), which is not what we want. For the region bounded by the curve and the x-axis, we must take the absolute value of the function, or equivalently, split the interval wherever the function changes sign and add the absolute areas.

For $y = \cos x$ between $0$ and $\pi$, the curve is positive from $0$ to $\pi/2$ and negative from $\pi/2$ to $\pi$. So the total area is the sum of the area above the axis and the area below the axis (taken as positive).

1. Find where the curve crosses the x-axis.

   $\cos x = 0$ at $x = \pi/2$ within $[0, \pi]$. This is the split point.

2. Area from $x = 0$ to $x = \pi/2$.

   Here $\cos x \ge 0$, so the area is simply the integral:

3. Area from $x = \pi/2$ to $x = \pi$. Here $\cos x \le 0$, so the area is the absolute value of the integral:

4. Total area.

Concept: area by integrating with respect to y (the region is bounded by the parabola on one side and the vertical line x = 1 on the other).

Curve: y^2 = 4a^2 (x - 1) => x = 1 + y^2/(4a^2). It is a rightward parabola with vertex (1, 0).

Boundaries: the line x = 1 (the tangent at the vertex) and the horizontal line y = 4a.

For a horizontal strip at height y (0 <= y <= 4a), the width is

x_curve - x_line = [1 + y^2/(4a^2)] - 1 = y^2/(4a^2).

Area = integral from 0 to 4a of y^2/(4a^2) dy

= (1/(4a^2)) * [y^3/3] from 0 to 4a

= (1/(4a^2)) * (64 a^3 / 3)

= 16a/3.

So the area is (16/3) a square units.

ANSWER: A

Step 1 — Check the sign first (the step students skip).

Area is $\displaystyle\int |y|\,dx$, so we must know where $y$ is negative. Here

and $\sin\theta\ge0$ for all $\theta\in[0,\pi]$. So $y=\sin\frac{x}{3}\ge0$ on the whole interval — the curve is one complete positive arch, sitting entirely above the $x$-axis. The modulus can therefore be dropped and the interval need not be split.

Step 2 — Set up the area integral.

Step 3 — Antiderivative.

Since $\displaystyle\int \sin(kx)\,dx=-\frac{1}{k}\cos(kx)$, with $k=\frac13$ the factor is $\frac{1}{k}=3$:

Step 4 — Evaluate.

Sanity check. For $y=\sin x$ one arch (from $0$ to $\pi$) has area $2$. Here the curve is horizontally stretched by a factor $3$ with the same height $1$, so the area should be $3\times2=6$. ✓ (Option (A) $9$ would need a further stretch; option (D) $3$ forgets the factor of $2$ from the arch.)

ANSWER: C

Concept & Intuition

The parabola $y^2 = 8x$ opens to the right. Its latus rectum is the vertical line through the focus. For $y^2 = 4ax$, the focus is at $(a,0)$ and the latus rectum is $x = a$. Here $4a = 8 \Rightarrow a = 2$, so the latus rectum is $x = 2$. The region bounded by the parabola, this vertical line, and the $x$-axis is the part of the parabola’s “bowl” from the vertex $(0,0)$ out to $x=2$, above the $x$-axis. Since the parabola is symmetric about the $x$-axis, the area above the axis is exactly half the total area between the curve and its latus rectum. We’ll integrate the top half ($y = \sqrt{8x}$) from $x=0$ to $x=2$.

Step-by-step solution

1. Identify the parabola’s parameters

   Standard form: $y^2 = 4ax$. Comparing with $y^2 = 8x$ gives $4a = 8 \Rightarrow a = 2$.

   Focus: $(2,0)$. Latus rectum: the line $x = 2$.

2. Set up the area integral

   The required region is bounded by:

   • the parabola $y^2 = 8x$ (upper branch: $y = \sqrt{8x}$),
   • the vertical line $x = 2$ (latus rectum),
   • the $x$-axis ($y = 0$). So area $A = \int_{x=0}^{2} y \, dx = \int_{0}^{2} \sqrt{8x} \, dx$.

3. Simplify the integrand

   $\sqrt{8x} = \sqrt{8}\sqrt{x} = 2\sqrt{2} \, x^{1/2}$.

4. Integrate

5. Evaluate $2^{3/2} = (2^3)^{1/2} = \sqrt{8} = 2\sqrt{2}$. So $A = \frac{4\sqrt{2}}{3} \cdot 2\sqrt{2} = \frac{4\sqrt{2} \cdot 2\sqrt{2}}{3} = \frac{8 \cdot 2}{3} = \frac{16}{3}$.

The key idea is that the latus rectum of a parabola is the chord through the focus perpendicular to the axis. For the standard parabola $y^2 = 4ax$, the focus is at $(a, 0)$ and the latus rectum is the vertical line $x = a$. The region bounded by the parabola and this line is symmetric about the x-axis, so we can compute the area in the upper half and double it.

Why this works: The parabola opens to the right, and the latus rectum acts as a vertical boundary. The area between the curve and this line from the vertex (at $x=0$) to the latus rectum is a classic application of integration with respect to $x$, using the fact that $y = \pm \sqrt{36x}$.

1. Identify the parameter $a$.

   The given parabola is $y^2 = 36x$. Compare with the standard form $y^2 = 4ax$.

   Here $4a = 36$, so $a = 9$.

   The focus is at $(9, 0)$, and the latus rectum is the line $x = 9$.

2. Set up the integral for the upper half.

   The upper branch of the parabola is $y = \sqrt{36x} = 6\sqrt{x}$.

   The region bounded by the parabola and the latus rectum in the first quadrant runs from $x = 0$ (vertex) to $x = 9$ (latus rectum).

   The area of this upper half is:

3. Evaluate the integral.

Evaluate from $0$ to $9$:

So the area of the upper half is $108$ square units.

4. Double to get the total area. Since the parabola is symmetric about the x-axis, the total area bounded by the parabola and its latus rectum is: