Let the area of the region \(x, y): x^2y + 4 ge 0, x+2y^2ge0, x+4y^2le8, yge0\$\{(x, y): x^2y + 4 \ge 0, x+2y^{2}\ge0, x+4y^{2}\le8, y\ge0\}$ be fracmn$\frac{m}{n}$ where m$m$ and n$n$ are coprime numbers. Then m+n$m+n$ is equal to:

### Related Formula textArea = int_y_1^y_2 (x_textright - x_textleft) dy$$\text{Area} = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) dy$$ ### Core Logic We need to find the area bounded by the curves in the first quadrant (since y ge 0$y \ge 0$). Note that the first inequality x^2y + 4 ge 0$x^2y + 4 \ge 0$ is trivially satisfied for all x, y ge 0$x, y \ge 0$. So we focus on bounding x$x$ using: Right curve: x = 8 - 4y^2$x = 8 - 4y^2$ Left curve: x = -2y^2$x = -2y^2$ 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 - 4y^2 = -2y^2 Rightarrow 2y^2 = 8 Rightarrow y^2 = 4 Rightarrow y = 2$8 - 4y^2 = -2y^2 \Rightarrow 2y^2 = 8 \Rightarrow y^2 = 4 \Rightarrow y = 2$ (Since y ge 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 x^2y + 4 ge 0$x^2y + 4 \ge 0$. If x = -2y^2$x = -2y^2$, then y$y$ must be bounded, but wait - the question limits are actually split into two regions depending on x^2y+4 ge 0$x^2y+4 \ge 0$? Wait, the first inequality x^2y+4 ge 0$x^2y+4 \ge 0$ might not be trivial if x$x$ is negative. If x = -2y^2$x = -2y^2$, then (-2y^2)^2 y + 4 ge 0 Rightarrow 4y^5 + 4 ge 0$(-2y^2)^2 y + 4 \ge 0 \Rightarrow 4y^5 + 4 \ge 0$, which is true for all y ge 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+2y^2 ge 0 Rightarrow x ge -2y^2$x+2y^2 \ge 0 \Rightarrow x \ge -2y^2$. Wait, the solution states another left curve: 2y-4$2y-4$. Is x^2y+4 ge 0$x^2y+4 \ge 0$ actually x + 2y - 4 ge 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 ge 0 Rightarrow x ge 2y - 4$x - 2y + 4 \ge 0 \Rightarrow x \ge 2y - 4$! Let's assume the left boundary splits between x = -2y^2$x = -2y^2$ and x = 2y - 4$x = 2y - 4$. ### Step 2: Region Bounds Intersection of x = -2y^2$x = -2y^2$ and x = 2y - 4$x = 2y - 4$: -2y^2 = 2y - 4 Rightarrow y^2 + y - 2 = 0 Rightarrow (y+2)(y-1) = 0 Rightarrow 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 - 4y^2$x = 8 - 4y^2$: 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$2y - 4 = 8 - 4y^2 \Rightarrow 4y^2 + 2y - 12 = 0 \Rightarrow 2y^2 + y - 6 = 0 \Rightarrow (2y-3)(y+2) = 0 \Rightarrow y = 3/2$. 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$y=1$): A_1 = int_0^1 ((8 - 4y^2) - (-2y^2)) dy = int_0^1 (8 - 2y^2) dy$$A_1 = \int_0^1 ((8 - 4y^2) - (-2y^2)) dy = \int_0^1 (8 - 2y^2) dy$$ A_1 = left[ 8y - frac2y^33 right]_0^1 = 8 - frac23 = frac223$$A_1 = \left[ 8y - \frac{2y^3}{3} \right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$ Region 2 (from y=1$y=1$ to y=3/2$y=3/2$): A_2 = int_1^3/2 ((8 - 4y^2) - (2y - 4)) dy = int_1^3/2 (12 - 2y - 4y^2) dy$$A_2 = \int_1^{3/2} ((8 - 4y^2) - (2y - 4)) dy = \int_1^{3/2} (12 - 2y - 4y^2) dy$$ A_2 = left[ 12y - y^2 - frac4y^33 right]_1^3/2$$A_2 = \left[ 12y - y^2 - \frac{4y^3}{3} \right]_1^{3/2}$$ A_2 = left( 12left(frac32right) - frac94 - frac43left(frac278right) right) - left( 12 - 1 - frac43 right)$$A_2 = \left( 12\left(\frac{3}{2}\right) - \frac{9}{4} - \frac{4}{3}\left(\frac{27}{8}\right) \right) - \left( 12 - 1 - \frac{4}{3} \right)$$ A_2 = left( 18 - frac94 - frac92 right) - left( 11 - frac43 right) = left( 18 - frac274 right) - frac293 = frac454 - frac293 = frac135 - 11612 = frac1912$$A_2 = \left( 18 - \frac{9}{4} - \frac{9}{2} \right) - \left( 11 - \frac{4}{3} \right) = \left( 18 - \frac{27}{4} \right) - \frac{29}{3} = \frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$ ### Step 4: Final Output Total Area A = A_1 + A_2$A = A_1 + A_2$: A = frac223 + frac1912 = frac88 + 1912 = frac10712$$A = \frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$ 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 ge 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Integrals