Let the area of the region (x, y): x²y + 4 ≥ 0, x+2y²≥0, x+4y²≤8, y≥0$\{(x, y): x^2y + 4 \ge 0, x+2y^{2}\ge0, x+4y^{2}\le8, y\ge0\}$ be (m)/(n)$\frac{m}{n}$ where m$m$ and n$n$ are coprime numbers. Then m+n$m+n$ is equal to:

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