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

Concept and Intuition

Area (as opposed to net signed integral) must always be non-negative. When $y=f(x)$ changes sign within $[a,b]$, the plain definite integral $\int_a^b f\,dx$ would let positive and negative parts cancel, understating the true enclosed area. So we must locate where $f$ crosses zero, integrate each sign-consistent piece separately, and add the absolute values.

Step-by-Step Solution

1. Factor: $x^2-5x+4=(x-1)(x-4)$, roots at $x=1,4$. Since the parabola opens upward, $y>0$ outside $[1,4]$ and $y<0$ inside $[1,4]$.
2. On $[0,2]$: $y>0$ on $[0,1)$ and $y<0$ on $(1,2]$.
3. Antiderivative: $F(x)=\dfrac{x^3}{3}-\dfrac{5x^2}{2}+4x$.
4. $F(0)=0$, $F(1)=\dfrac13-\dfrac52+4=\dfrac{2-15+24}{6}=\dfrac{11}{6}$, $F(2)=\dfrac83-10+8=\dfrac83-2=\dfrac23$.
5. $\int_0^1y\,dx=F(1)-F(0)=\dfrac{11}{6}$ (positive region, keep as is).
6. $\int_1^2y\,dx=F(2)-F(1)=\dfrac23-\dfrac{11}{6}=\dfrac{4-11}{6}=-\dfrac76$ (negative region, take absolute value $\dfrac76$).
7. Total area $=\dfrac{11}{6}+\dfrac76=\dfrac{18}{6}=3$.

Common Mistakes

• Directly computing $\int_0^2y\,dx$ (net signed area) without splitting at $x=1$, which gives $\dfrac{11}{6}-\dfrac76=\dfrac23$ — one of the wrong-looking distractor values, not the true area.
• Forgetting that the region below the axis contributes a positive area, not a negative one.

ANSWER: B

Concept and Intuition

"Area bounded by a curve and the x-axis" always means the geometric (unsigned) area, not the signed definite integral. Whenever the curve crosses the x-axis within the given range, you must split the integral at the crossing point(s) and take the absolute value of each piece separately, because a naive single integral would let a below-axis region cancel an above-axis region.

Step-by-Step Solution

1. Find where $y=x^2-4$ crosses the x-axis: $x^2-4=0 \Rightarrow x=\pm2$. Both roots lie within $[-2,3]$, splitting the region into $[-2,2]$ (curve below axis, since $x^2<4$ there) and $[2,3]$ (curve above axis).
2. The antiderivative is $F(x)=\dfrac{x^3}{3}-4x$.
3. Piece 1, $[-2,2]$:

This is negative (curve below axis), so the area contributed is $\left|-\frac{32}{3}\right|=\frac{32}{3}$.

4. Piece 2, $[2,3]$:

This is positive (curve above axis), so the area contributed is $\frac{7}{3}$.

5. Total area $=\dfrac{32}{3}+\dfrac{7}{3}=\dfrac{39}{3}=13$.

Common Mistakes

• Integrating $\int_{-2}^{3}(x^2-4)\,dx$ directly as a single signed integral, which lets the negative and positive regions partially cancel and gives the wrong (smaller) number.
• Forgetting that $x=-2$ is also a root and treating $[-2,2]$ as entirely "extra" rather than as its own below-axis region needing an absolute value.

ANSWER: A

Concept and Intuition

Both curves are upward-opening parabolas with vertices on the X-axis at $x=0$ and $x=6$. They cross exactly once between these vertices. The bounded region is a "leaf" shape: from $x=0$ to $x=3$ it's capped by the first parabola (rising from $0$ to $1$), and from $x=3$ to $x=6$ it's capped by the second parabola (falling from $1$ back to $0$), with the X-axis as the base throughout.

Step-by-Step Solution

1. Write both parabolas as functions of $x$:

2. Find their intersection: $x^2=(x-6)^2 \Rightarrow x^2 = x^2-12x+36 \Rightarrow 12x=36 \Rightarrow x=3$. At $x=3$: $y=9/9=1$. So they meet at $(3,1)$.

3. Determine which curve is the "upper boundary" on each side:

   • For $0\le x\le3$: $y_1=x^2/9$ rises from $0$ to $1$; $y_2=(x-6)^2/9$ falls from $4$ to $1$ — so $y_1 < y_2$ here, meaning the region's upper boundary (the curve closer to the axis, bounding the smaller enclosed area) is $y_1$.
   • For $3\le x\le6$: by symmetry, $y_2$ is now the smaller one, forming the boundary.

   So the bounded "leaf" region between the axis and the lower envelope of the two parabolas has area:

4. Compute each piece:

5. Total area $=1+1=2$.

Common Mistakes

• Integrating the wrong curve over the wrong sub-interval (mixing up which parabola dominates on which side of $x=3$).
• Forgetting the substitution shift when integrating $(x-6)^2$, leading to arithmetic errors in the limits.

ANSWER: C

Concept and Intuition

$y^2=8(x+2)$ is a rightward-opening parabola with vertex $(-2,0)$; $y^2=4(1-x)$ is a leftward-opening parabola with vertex $(1,0)$. They cross where $8(x+2)=4(1-x)$. Between that crossing point and the $y$-axis, one parabola sits outside the other, so the area enclosed by both curves together with the line $x=0$ is the region trapped between them over that stretch, doubled for the symmetric halves above and below the $x$-axis.

Step-by-Step Solution

1. Find the intersection: $8(x+2)=4(1-x)\Rightarrow 8x+16=4-4x\Rightarrow 12x=-12\Rightarrow x=-1$, giving $y^2=8(1)=8\Rightarrow y=\pm2\sqrt2$.
2. At $x=0$: first curve gives $y^2=16\Rightarrow y=\pm4$; second curve gives $y^2=4\Rightarrow y=\pm2$. So on $[-1,0]$, curve 1 ($y^2=8(x+2)$) is the outer boundary and curve 2 ($y^2=4(1-x)$) is the inner one.
3. Area of the upper half between the curves: $A_{upper}=\displaystyle\int_{-1}^{0}\Big[\sqrt{8(x+2)}-\sqrt{4(1-x)}\Big]dx$.
4. $\displaystyle\int_{-1}^{0}\sqrt{8(x+2)}\,dx$: substitute $u=x+2$ ($u:1\to2$): $=\sqrt8\left[\dfrac23 u^{3/2}\right]_1^2=2\sqrt2\cdot\dfrac23\left(2\sqrt2-1\right)=\dfrac{16-4\sqrt2}{3}$.
5. $\displaystyle\int_{-1}^{0}\sqrt{4(1-x)}\,dx=2\displaystyle\int_{-1}^{0}\sqrt{1-x}\,dx$: substitute $v=1-x$ ($v:2\to1$): $=2\left[\dfrac23 v^{3/2}\right]_1^2\text{(reversed)}=\dfrac{8\sqrt2-4}{3}$.
6. $A_{upper}=\dfrac{16-4\sqrt2}{3}-\dfrac{8\sqrt2-4}{3}=\dfrac{20-12\sqrt2}{3}$.
7. Total area (doubling for the lower half by symmetry): $A=2\times\dfrac{20-12\sqrt2}{3}=\dfrac{40-24\sqrt2}{3}=\dfrac83(5-3\sqrt2)$.

Common Mistakes

• Integrating over the wrong interval — including the region from $x=-2$ to $-1$ (which lies to the left of the intersection and is NOT bounded by the $y$-axis condition).
• Swapping which curve is outer vs. inner on $[-1,0]$, giving a negative or wrong-magnitude area.

ANSWER: A

Concept and Intuition

$y=x^2$ opens upward from the origin, while $y=8-x^2$ opens downward from $(0,8)$ — together they bound a lens-shaped (vesica-like) region symmetric about the $y$-axis. The area between two curves $y=f(x)$ (upper) and $y=g(x)$ (lower) over $[a,b]$ is $\int_a^b[f(x)-g(x)]dx$, where $[a,b]$ are the intersection points.

Step-by-Step Solution

1. Find intersections: set $x^2=8-x^2 \Rightarrow 2x^2=8 \Rightarrow x^2=4 \Rightarrow x=\pm2$.
2. For $-2<x<2$: check which curve is on top — at $x=0$, $y=x^2=0$ and $y=8-x^2=8$, so $8-x^2$ is the upper curve.
3. Area $=\displaystyle\int_{-2}^{2}\left[(8-x^2)-x^2\right]dx=\int_{-2}^{2}(8-2x^2)\,dx$.
4. The integrand is even, so $=2\displaystyle\int_0^2(8-2x^2)\,dx$.
5. $\displaystyle\int_0^2(8-2x^2)\,dx=\left[8x-\frac{2x^3}{3}\right]_0^2=8(2)-\frac{2(8)}{3}=16-\frac{16}{3}=\frac{48-16}{3}=\frac{32}{3}$.
6. Area $=2\times\dfrac{32}{3}=\dfrac{64}{3}$ sq. units.

Common Mistakes

• Forgetting to double the integral over $[0,2]$ (or equivalently integrating over the wrong interval).
• Mixing up which parabola is "on top" — a quick check at $x=0$ resolves it instantly.
• Sign errors subtracting $x^2$ from $8-x^2$.

ANSWER: C

Concept and Intuition

The area between two curves over an interval is $\int (\text{upper}-\text{lower})\,dx$, where the interval's endpoints are found by setting the two expressions equal (their intersection points).

Step-by-Step Solution

1. Find intersection: $2x-x^2 = -x \Rightarrow 2x-x^2+x=0 \Rightarrow 3x-x^2=0 \Rightarrow x(3-x)=0 \Rightarrow x=0$ or $x=3$.
2. Check which curve is on top between $0$ and $3$: at $x=1$, parabola gives $2(1)-1=1$; line gives $-1$. Parabola is above.
3. Area $= \displaystyle\int_0^3\left[(2x-x^2)-(-x)\right]dx = \int_0^3(3x-x^2)\,dx$.
4. $= \left[\dfrac{3x^2}{2}-\dfrac{x^3}{3}\right]_0^3 = \dfrac{3(9)}{2}-\dfrac{27}{3} = \dfrac{27}{2}-9 = \dfrac{27-18}{2}=\dfrac{9}{2}$.

Common Mistakes

• Forgetting to add $x$ back (i.e., writing the integrand as $2x-x^2+x$ correctly, not $2x-x^2-x$).
• Not checking which curve is on top and getting a negative area / wrong sign.

ANSWER: D

Concept and Intuition

When a region is described by $g(y) \le x \le f(y)$, its area is $\int (f(y)-g(y))\,dy$ taken between the $y$-values where the two boundary curves meet. Here $g(y)=y^2/2$ (parabola opening rightward) and $f(y)=y+4$ (line).

Step-by-Step Solution

1. Find intersections: $\dfrac{y^2}{2} = y+4 \Rightarrow y^2 = 2y+8 \Rightarrow y^2-2y-8=0 \Rightarrow (y-4)(y+2)=0$, so $y=-2$ and $y=4$.
2. Between these, the line is to the right of the parabola, so width $= (y+4)-\dfrac{y^2}{2}$.
3. Area $= \displaystyle\int_{-2}^{4}\left(y+4-\dfrac{y^2}{2}\right)dy = \left[\dfrac{y^2}{2}+4y-\dfrac{y^3}{6}\right]_{-2}^{4}$.
4. At $y=4$: $8+16-\dfrac{64}{6}=24-\dfrac{32}{3}=\dfrac{40}{3}$. At $y=-2$: $2-8+\dfrac{8}{6}=-6+\dfrac{4}{3}=-\dfrac{14}{3}$.
5. Area $= \dfrac{40}{3}-\left(-\dfrac{14}{3}\right)=\dfrac{54}{3}=18$.

Common Mistakes

• Integrating with respect to $x$ and mishandling the two-valued parabola instead of integrating cleanly in $y$.
• Dropping the sign when evaluating the antiderivative at the lower limit $y=-2$.

ANSWER: B

Concept and Intuition

The area between a curve and a line over an interval is the integral of (upper function − lower function). First find the intersection points to get the limits, then check which function is on top between them.

Step-by-Step Solution

1. Rewrite the line: $7x-y+1=0 \Rightarrow y=7x+1$.
2. Find intersections with $y=x^2+5x+1$:

3. Check which curve is on top for $x\in(0,2)$, e.g. at $x=1$: line gives $8$, parabola gives $1+5+1=7$. So the line is above the parabola.
4. Area:

5. $=\left[x^2-\frac{x^3}{3}\right]_0^2=(4-\tfrac83)-0=\frac{4}{3}$

Common Mistakes

• Subtracting in the wrong order (parabola minus line), giving a negative area instead of taking the correct (upper − lower) order.
• Arithmetic slip computing $4-8/3$.

ANSWER: C

Concept and Intuition

"Inside the circle, outside the parabola" in the first quadrant means, for each height $y$, taking the horizontal strip between $x=0$ (or the parabola, whichever is the boundary of "outside") out to the circle — but only where that strip is genuinely outside the parabola. The curves meet where the circle passes through the latus-rectum end, so the region naturally splits into two ranges of $y$.

Step-by-Step Solution

1. $y^2=12x\Rightarrow 4a=12\Rightarrow a=3$. Latus rectum ends: $(a,\pm2a)=(3,\pm6)$.
2. Circle centred at vertex (origin) through $(3,6)$: $r^2=3^2+6^2=45$, i.e. $x^2+y^2=45$.
3. For $0\le y\le6$: parabola boundary is $x_p(y)=y^2/12$ and circle boundary is $x_c(y)=\sqrt{45-y^2}$; checking, $x_p(y)\le x_c(y)$ throughout (equal at $y=6$), so the "outside parabola" strip $0\le x\le x_p(y)$ lies wholly inside the circle. Its area is $\displaystyle\int_0^6\frac{y^2}{12}dy=\frac{1}{12}\cdot\frac{6^3}{3}=6$.
4. For $6\le y\le\sqrt{45}$: now $x_c(y)<x_p(y)$, so the entire circular cross-section $0\le x\le x_c(y)$ is outside the parabola. Its area is $\displaystyle\int_6^{\sqrt{45}}\sqrt{45-y^2}\,dy$.
5. Using $\int\sqrt{a^2-y^2}dy=\frac y2\sqrt{a^2-y^2}+\frac{a^2}2\mathrm{Sin}^{-1}\!\left(\frac ya\right)$ with $a^2=45$: at $y=\sqrt{45}$, value $=\frac{45}{2}\cdot\frac\pi2=\frac{45\pi}4$; at $y=6$, value $=9+\frac{45}{2}\mathrm{Sin}^{-1}\!\left(\frac{2}{\sqrt5}\right)$ (since $6/\sqrt{45}=2/\sqrt5$).
6. So this piece $=\frac{45\pi}{4}-9-\frac{45}{2}\mathrm{Sin}^{-1}\!\left(\frac2{\sqrt5}\right)$.
7. Total area $=6+\frac{45\pi}4-9-\frac{45}2\mathrm{Sin}^{-1}\!\left(\frac2{\sqrt5}\right)=-3+\frac{45\pi}4-\frac{45}2\mathrm{Sin}^{-1}\!\left(\frac{2}{\sqrt5}\right)$.
8. Since $\frac2{\sqrt5}$ and $\frac1{\sqrt5}$ are complementary sines ($\mathrm{Sin}^{-1}(2/\sqrt5)=\pi/2-\mathrm{Sin}^{-1}(1/\sqrt5)$), substitute: total $=-3+\frac{45\pi}4-\frac{45}2\left(\frac\pi2-\mathrm{Sin}^{-1}\!\left(\frac1{\sqrt5}\right)\right)=-3+\frac{45}{2}\mathrm{Sin}^{-1}\!\left(\frac1{\sqrt5}\right)$ (the $\pi$ terms cancel).

Common Mistakes

• Treating the whole quarter-disk cross-section as "outside the parabola" for all $y$ (only true for $y>6$; for $y<6$ only the strip up to the parabola counts).
• Sign/complementary-angle slip converting $\mathrm{Sin}^{-1}(2/\sqrt5)$ to $\mathrm{Sin}^{-1}(1/\sqrt5)$.

ANSWER: A