The area (in sq. units) of the region (x, y): 0 ≤ y ≤ 2|x| + 1, 0 ≤ y ≤ x² + 1, |x| ≤ 3$$\{(x, y): 0 \le y \le 2|x| + 1, 0 \le y \le x^2 + 1, |x| \le 3\}$$ is

Related Formula

Core Logic

For A₁$A_1$: Intersection of y = x² + 2$y = x^2 + 2$ and x + y = 8$x + y = 8$. x + (x² + 2) = 8 ⇒ x² + x - 6 = 0 ⇒ (x+3)(x-2) = 0$x + (x^2 + 2) = 8 \Rightarrow x^2 + x - 6 = 0 \Rightarrow (x+3)(x-2) = 0$. In first quadrant, x = 2$x = 2$. Intersection point is (2, 6)$(2, 6)$.

Step 1: Calculate A1

Step 2: Area A2 Definition

For A₂$A_2$: Enclosed by y = x² + 2$y = x^2 + 2$, y = √(x)$y = \sqrt{x}$, x=2$x=2$ and y-axis. Upper curve: y = x²+2$y = x^2+2$, Lower curve: y = √(x)$y = \sqrt{x}$.

Step 3: Calculate A2

Step 4: Difference of Areas

Pattern Recognition

Carefully identify the bounding regions. A₁$A_1$ uses a line as the upper bound, A₂$A_2$ strips away a root curve as its lower bound. Integrating separately before subtracting avoids algebraic alignment errors.

Chapter Mix

Class 12 Maths: Application of Integrals

Core Logic

• y ≥ 1$y \geq 1$ (horizontal line)
• y ≤ x²$y \leq x^2$ (upward-opening parabola)
• xy ≤ 8 y ≤ (8)/(x)$xy \leq 8 \implies y \leq \frac{8}{x}$ (rectangular hyperbola)

Find points of intersection: Intersection of y = x²$y = x^2$ and y = (8)/(x)$y = \frac{8}{x}$:

At x = 2$x = 2$, y = 4$y = 4$. So they meet at (2, 4)$(2, 4)$.

Intersection of y = x²$y = x^2$ and y = 1$y = 1$:

Intersection of y = (8)/(x)$y = \frac{8}{x}$ and y = 1$y = 1$:

The region spans from x = 1$x = 1$ to x = 8$x = 8$. The upper boundary changes at x = 2$x = 2$.

Step 1: Set up Integrals

For x in [1, 2]$x \in [1, 2]$, the upper curve is y = x²$y = x^2$ and the lower is y = 1$y = 1$. For x in [2, 8]$x \in [2, 8]$, the upper curve is y = (8)/(x)$y = \frac{8}{x}$ and the lower is y = 1$y = 1$.

Step 2: Integration

Calculate the first integral:

Calculate the second integral:

Since ln 8 = 3 ln 2$\ln 8 = 3 \ln 2$:

Total Area A$A$:

Chapter Mix

Class 12 Mathematics: Application of Integrals

Related Formula

Area enclosed by y = f(x)$y = f(x)$ and y = g(x)$y = g(x)$ from x=a$x=a$ to x=b$x=b$:

Area between x² = 4ay$x^2 = 4ay$ and line x = my$x = my$ is (8a²)/(3m³)$\frac{8a^2}{3m^3}$.

Core Logic

Find Points of Intersection for P₁$P_1$ and P₂$P_2$:

Area bounded between P₁$P_1$ and P₂$P_2$:

Execution

Area between P₁$P_1$ and y = α x$y = \alpha x$ is given as 108 / 6 = 18$108 / 6 = 18$ sq. units. For x² = 4ay$x^2 = 4ay$ and x = my$x = my$, area is (8a²)/(3m³)$\frac{8a^2}{3m^3}$. Here, the parabola is x² = (y)/(4)$x^2 = \frac{y}{4}$ and line is x = (y)/(α)$x = \frac{y}{\alpha}$. Thus, 4a = (1)/(4) ⇒ a = (1)/(16)$4a = \frac{1}{4} \Rightarrow a = \frac{1}{16}$, and m = (1)/(α)$m = \frac{1}{\alpha}$.

Applying the formula:

α = 12$\alpha = 12$

Pattern Recognition

Using the standard result (8a²)/(3m³)$\frac{8a^2}{3m^3}$ for the area bounded between x²=4ay$x^2=4ay$ and x=my$x=my$ avoids tedious integration for straight lines intersecting parabolas at the origin.

Chapter Mix

Class 12 Maths: Application of Integrals

Related Formula

Area bounded by functions integrated with respect to y$y$ axis:

Core Logic

Identify the bounding graphs and intersection coordinates in the first quadrant, then construct standard definite integrals along the vertical axis.

Step 1: Unpack Bounding Curves

The condition |4-x²| ≤ y$|4-x^2| \le y$ splits into two sections at x=2$x=2$:

• For 0 ≤ x ≤ 2 4 - x² ≤ y x² ≥ 4 - y x = √(4-y)$0 \le x \le 2 \implies 4 - x^2 \le y \implies x^2 \ge 4 - y \implies x = \sqrt{4-y}$
• For x ≥ 2 x² - 4 ≤ y x² ≤ 4 + y x = √(4+y)$x \ge 2 \implies x^2 - 4 \le y \implies x^2 \le 4 + y \implies x = \sqrt{4+y}$

Also bounded by y ≤ x² x ≥ √(y)$y \le x^2 \implies x \ge \sqrt{y}$, and the outer cap constraint y ≤ 4$y \le 4$.

Step 2: Construct the Integral Area Formula

Integrating with respect to y$y$ covers the region bounded on the left by √(y)$\sqrt{y}$ and √(4-y)$\sqrt{4-y}$, and on the right by √(4+y)$\sqrt{4+y}$:

Step 3: Evaluate the Definite Integrals

Evaluating these values precisely:

Step 4: Solve for Constants

Compare the final value expression to ( 80√(2)α - β)$\left(\frac{80\sqrt{2}}{\alpha} - \beta\right)$:

Pattern Recognition

Integrating along the vertical axis (y$y$-direction) is significantly faster here because it avoids splitting the domain across multiple vertical segments on the horizontal x$x$-axis.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Related Formula

The area between two continuous curves yupper(x)$y_{\text{upper}}(x)$ and ylower(x)$y_{\text{lower}}(x)$ from x=a$x=a$ to x=b$x=b$ is given by:

Core Logic

Analyze the given inequality condition |x - y| ≤ y ≤ 4√(x)$|x - y| \le y \le 4\sqrt{x}$:

• Left Inequality:

Adding y$y$ across all terms: 0 ≤ x ≤ 2y$0 \le x \le 2y$ This yields two conditions:

• x ≥ 0$x \ge 0$
• y ≥ (x)/(2)$y \ge \frac{x}{2}$ (lower bounding line)
• Right Inequality:

This gives the upper bounding parabolic curve y = 4√(x)$y = 4\sqrt{x}$.

Thus, the enclosed region lies in the first quadrant bounded between the upper curve y = 4√(x)$y = 4\sqrt{x}$ and the lower line y = (x)/(2)$y = \frac{x}{2}$.

Step 1: Finding Points of Intersection

Equating the two boundary curves:

Corresponding y$y$-values:

• At x = 0 y = 0$x = 0 \implies y = 0$
• At x = 64 y = 32$x = 64 \implies y = 32$

The curves intersect at (0, 0)$(0, 0)$ and (64, 32)$(64, 32)$.

Step 2: Area Integration

Integrating with respect to x$x$ from x = 0$x = 0$ to x = 64$x = 64$: aligned Area &= ∫₀⁶⁴ ( 4√(x) - (x)/(2) ) dx &= [ (8)/(3)x3/2 - (x²)/(4) ]₀⁶⁴ &= (8)/(3)(64)3/2 - ((64)²)/(4) &= (8)/(3)(512) - (4096)/(4) &= (4096)/(3) - 1024 &= (4096 - 3072)/(3) &= (1024)/(3) aligned$$ \begin{aligned} \text{Area} &= \int_0^{64} \left( 4\sqrt{x} - \frac{x}{2} \right) \, dx \\ &= \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{4} \right]_0^{64} \\ &= \frac{8}{3}(64)^{3/2} - \frac{(64)^2}{4} \\ &= \frac{8}{3}(512) - \frac{4096}{4} \\ &= \frac{4096}{3} - 1024 \\ &= \frac{4096 - 3072}{3} \\ &= \frac{1024}{3} \end{aligned} $$

Pattern Recognition (Alternative Verification)

For the area enclosed between a parabola y² = 4ax$y^2 = 4ax$ and a line y = mx$y = mx$:

Given y² = 16x 4a = 16 a = 4$y^2 = 16x \implies 4a = 16 \implies a = 4$, and y = (1)/(2)x m = (1)/(2)$y = \frac{1}{2}x \implies m = \frac{1}{2}$:

Evaluation Rubric / Model Answer

Correct Answer: Option B ((1024)/(3)$\frac{1024}{3}$)

Chapter Mix

• Class 12 Mathematics: Applications of Integrals
• Class 11 Mathematics: Conic Sections