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

Let us find the point of intersection of the curves y = x² + 2$y = x^2 + 2$ and y = 2x + 2$y = 2x + 2$:

Evaluating behaviors over boundaries:

• For 0 ≤ x ≤ 2$0 \leq x \leq 2$: x² + 2 ≤ 2x + 2 = x² + 2$x^2 + 2 \leq 2x + 2 \implies \min = x^2 + 2$
• For 2 ≤ x ≤ 3$2 \leq x \leq 3$: 2x + 2 ≤ x² + 2 = 2x + 2$2x + 2 \leq x^2 + 2 \implies \min = 2x + 2$

Step 1: Integration Resolution

Setting up continuous area integral steps:

Step 2: Scaling the Output Value

We need to compute 12A$12A$:

Pattern Recognition

For min/max boundary sets, always compute intersections first to accurately split integration domains into separate regions.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Related Formula

Core Logic

We need to find the area bounded by the curves in the first quadrant (since y ≥ 0$y \ge 0$). Note that the first inequality x²y + 4 ≥ 0$x^2y + 4 \ge 0$ is trivially satisfied for all x, y ≥ 0$x, y \ge 0$. So we focus on bounding x$x$ using: Right curve: x = 8 - 4y²$x = 8 - 4y^2$ Left curve: x = -2y²$x = -2y^2$ However, there might be constraints where these cross or hit the axes. We must inspect intersections.

Step 1: Finding Intersection Points

Where do the left and right parabolas intersect? 8 - 4y² = -2y² ⇒ 2y² = 8 ⇒ y² = 4 ⇒ y = 2$8 - 4y^2 = -2y^2 \Rightarrow 2y^2 = 8 \Rightarrow y^2 = 4 \Rightarrow y = 2$ (Since y ≥ 0$y \ge 0$). Wait, does x$x$ have boundaries? Since x$x$ can't drop arbitrarily into negative territory if it's restricted by other axes. Let's check x²y + 4 ≥ 0$x^2y + 4 \ge 0$. If x = -2y²$x = -2y^2$, then y$y$ must be bounded, but wait - the question limits are actually split into two regions depending on x²y+4 ≥ 0$x^2y+4 \ge 0$? Wait, the first inequality x²y+4 ≥ 0$x^2y+4 \ge 0$ might not be trivial if x$x$ is negative. If x = -2y²$x = -2y^2$, then (-2y²)² y + 4 ≥ 0 ⇒ 4y⁵ + 4 ≥ 0$(-2y^2)^2 y + 4 \ge 0 \Rightarrow 4y^5 + 4 \ge 0$, which is true for all y ≥ 0$y \ge 0$. Wait, there seems to be a misinterpretation of the first inequality. Let's look closer at the PDF solution boundaries. The integration is broken at y=1$y=1$. Why? x+2y² ≥ 0 ⇒ x ≥ -2y²$x+2y^2 \ge 0 \Rightarrow x \ge -2y^2$. Wait, the solution states another left curve: 2y-4$2y-4$. Is x²y+4 ≥ 0$x^2y+4 \ge 0$ actually x + 2y - 4 ≥ 0$x + 2y - 4 \ge 0$? Yes, OCR shows `x^2y+4` but the solution integrates `(2y-4)`. Thus the original condition is likely x - 2y + 4 ≥ 0 ⇒ x ≥ 2y - 4$x - 2y + 4 \ge 0 \Rightarrow x \ge 2y - 4$! Let's assume the left boundary splits between x = -2y²$x = -2y^2$ and x = 2y - 4$x = 2y - 4$.

Step 2: Region Bounds

Intersection of x = -2y²$x = -2y^2$ and x = 2y - 4$x = 2y - 4$: -2y² = 2y - 4 ⇒ y² + y - 2 = 0 ⇒ (y+2)(y-1) = 0 ⇒ y = 1$-2y^2 = 2y - 4 \Rightarrow y^2 + y - 2 = 0 \Rightarrow (y+2)(y-1) = 0 \Rightarrow y = 1$. Intersection of x = 2y - 4$x = 2y - 4$ and x = 8 - 4y²$x = 8 - 4y^2$: 2y - 4 = 8 - 4y² ⇒ 4y² + 2y - 12 = 0 ⇒ 2y² + y - 6 = 0 ⇒ (2y-3)(y+2) = 0 ⇒ y = 3/2$2y - 4 = 8 - 4y^2 \Rightarrow 4y^2 + 2y - 12 = 0 \Rightarrow 2y^2 + y - 6 = 0 \Rightarrow (2y-3)(y+2) = 0 \Rightarrow y = 3/2$. So the region shifts left boundary at y=1$y=1$ and closes entirely at y=3/2$y=3/2$.

Step 3: Setting up the Integration

Region 1 (from y=0$y=0$ to y=1$y=1$):

Region 2 (from y=1$y=1$ to y=3/2$y=3/2$):

Step 4: Final Output

Total Area A = A₁ + A₂$A = A_1 + A_2$:

This implies m = 107$m = 107$ and n = 12$n = 12$. Since 107 and 12 are coprime, m + n = 107 + 12 = 119$m + n = 107 + 12 = 119$.

Pattern Recognition

When dealing with multiple inequalities bounded by y ≥ 0$y \ge 0$, always project horizontally (integrate wrt y$y$) as the bounds natively trace left-to-right distances. Find intersection nodes to partition the integral correctly.

Chapter Mix

Class 12 Maths: Application of Integrals

Related Formula

Core Logic

First, find the points of intersection between the circle x²+y²=169$x^2+y^2=169$ and the line 5x-y=13 ⇒ y = 5x-13$5x-y=13 \Rightarrow y = 5x-13$. Substitute y$y$ into the circle equation:

The solutions are x=0$x=0$ and x=5$x=5$. When x=0, y=-13$x=0, y=-13$. Point is (0, -13)$(0, -13)$. When x=5, y=12$x=5, y=12$. Point is (5, 12)$(5, 12)$.

The required area is bounded below the line x = (y+13)/(5)$x = \frac{y+13}{5}$ and above the right-hand boundary of the circle x = √(169-y²)$x = \sqrt{169-y^2}$ across the y-axis boundaries [-13, 12]$[-13, 12]$.

Step 1: Setup Area Integral

Integrate with respect to y$y$ (from left to right curves, bounded horizontally):

Split the integral into two parts: Part A (Circle): ∫₋₁₃¹² √(169-y²) dy$\int_{-13}^{12} \sqrt{169-y^2} dy$ Part B (Line): ∫₋₁₃¹² (y+13)/(5) dy$\int_{-13}^{12} \frac{y+13}{5} dy$

Step 2: Evaluate Integrals

Part A (Circle Integral):

Evaluate at upper limit 12:

Evaluate at lower limit -13:

Value of Part A = 30 + (169π)/(4) + (169)/(2) ⁻¹((12)/(13))$30 + \frac{169\pi}{4} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$

Part B (Line Integral - matches the area of the bounded triangle geometric region):

Step 3: Map to Requested Format

Subtract Part B from Part A:

Comparing this exactly with the given format (πα)/(2β) - (65)/(2) + (α)/(β) ⁻¹((12)/(13))$\frac{\pi\alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$: We see that (α)/(β) = (169)/(2)$\frac{\alpha}{\beta} = \frac{169}{2}$. Since 169 and 2 are coprime, α = 169$\alpha = 169$ and β = 2$\beta = 2$.

Calculate α + β$\alpha + \beta$: 169 + 2 = 171$169 + 2 = 171$

Pattern Recognition

When evaluating line integrals forming a triangle with horizontal bounds, bypass algebraic integration and visually calculate (1)/(2) · b · h$\frac{1}{2} \cdot b \cdot h$. Here, base=25 along y-axis, height=5 along x-axis, area = 125/2$125/2$. Instantly saves integration time.

Chapter Mix

Class 12 Mathematics: Application of Integrals Class 11 Mathematics: Straight Lines

Related Formula

Core Logic

To simplify calculations, shift the origin. Let X = x - 2$X = x - 2$. The equations become: Parabola: y² = 4X ⇒ X = (y²)/(4)$y^2 = 4X \Rightarrow X = \frac{y^2}{4}$ Line: y = 2(X + 2) - 8 ⇒ y = 2X - 4 ⇒ X = (y + 4)/(2)$y = 2(X + 2) - 8 \Rightarrow y = 2X - 4 \Rightarrow X = \frac{y + 4}{2}$

Step 1: Finding points of intersection

Set the X$X$ values equal to find intersection points in terms of y$y$:

y² = 2y + 8$y^2 = 2y + 8$

The intersection points are at y = -2$y = -2$ and y = 4$y = 4$.

Step 2: Area Integration

Integrate with respect to y$y$ from -2$-2$ to 4$4$:

Upper limit (y=4)$(y=4)$: (16)/(4) + 8 - (64)/(12) = 4 + 8 - (16)/(3) = 12 - (16)/(3) = (20)/(3)$\frac{16}{4} + 8 - \frac{64}{12} = 4 + 8 - \frac{16}{3} = 12 - \frac{16}{3} = \frac{20}{3}$ Lower limit (y=-2)$(y=-2)$: (4)/(4) - 4 - (-8)/(12) = 1 - 4 + (2)/(3) = -3 + (2)/(3) = -(7)/(3)$\frac{4}{4} - 4 - \frac{-8}{12} = 1 - 4 + \frac{2}{3} = -3 + \frac{2}{3} = -\frac{7}{3}$

The solution simplifies it directly to 9$9$ square units.

Pattern Recognition

For a horizontal parabola interacting with a line, integrating along the y-axis is always cleaner than splitting it into multiple integrals along the x-axis.

Chapter Mix

Class 12 Maths: Application of Integrals

Related Formula

Core Logic

Find intersection points of y = 4x - x²$y = 4x - x^2$ and 3y = (x - 4)²$3y = (x - 4)^2$:

Roots are x = 1, 4$x = 1, 4$.

Area integral:

Chapter Mix

Class 12 Maths: Applications of the Integrals