Let A₁$A_{1}$ be the bounded area enclosed by the curves y = x² + 2$y = x^{2} + 2$, x + y = 8$x + y = 8$ and y-axis that lies in the first quadrant. Let A₂$A_{2}$ be the bounded area enclosed by the curves y = x² + 2$y = x^{2} + 2$, y² = x$y^{2} = x$, x = 2$x = 2$, and y-axis that lies in the first quadrant. Then A₁ - A₂$A_{1} - A_{2}$ is equal to

Related Formula

Core Logic

Evaluate f(0)$f(0)$ by setting α = 0$\alpha = 0$ in the function for y$y$.

At α = 0 y = |0· x - 5| - |1 - 0· x| + 0· x²$\alpha = 0 \implies y = |0\cdot x - 5| - |1 - 0\cdot x| + 0\cdot x^2$

So for f(0)$f(0)$, the bounded area A₁$A_1$ is between y = 4$y = 4$, y = √(x)$y = \sqrt{x}$, from x = 0$x = 0$ to x = 1$x = 1$.

Step 1: Calculating f(0)

So, f(0) = (10)/(3)$f(0) = \frac{10}{3}$.

Step 2: Determining curve for f(1)

Evaluate f(1)$f(1)$ by setting α = 1$\alpha = 1$ in the function for y$y$.

At α = 1 y = |x - 5| - |1 - x| + x²$\alpha = 1 \implies y = |x - 5| - |1 - x| + x^2$. Since the area is restricted to x in [0, 1]$x \in [0, 1]$: x - 5$x - 5$ is negative, so |x - 5| = 5 - x$|x - 5| = 5 - x$. 1 - x$1 - x$ is positive, so |1 - x| = 1 - x$|1 - x| = 1 - x$.

Step 3: Calculating f(1)

For f(1)$f(1)$, the bounded area A₂$A_2$ is between y = 4 + x²$y = 4 + x^2$ and y = √(x)$y = \sqrt{x}$, from x = 0$x = 0$ to x = 1$x = 1$.

So, f(1) = (11)/(3)$f(1) = \frac{11}{3}$.

Step 4: Final Sum

Pattern Recognition

Modulus functions combined with parametric limits should be instantly resolved by plugging in the bounds of x$x$ (here [0,1]$[0,1]$) to drop the absolute value signs before attempting any integral.

Chapter Mix

Class 12 Maths: Area Under the Curve Class 11 Maths: Modulus Function

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