### Related Formula Method of algebraic parameter substitution: Set 3-x^2 = t^2 implies -2xmathrmdx = 2tmathrmdt implies xmathrmdx = -tmathrmdt$3-x^2 = t^2 \implies -2x\mathrm{d}x = 2t\mathrm{d}t \implies x\mathrm{d}x = -t\mathrm{d}t$ ### Core Logic Perform the specified variable parameter replacement steps [cite: 1361, 1362]: 3 - x^2 = t^2 implies x \, mathrmdx = -t \, mathrmdt$3 - x^2 = t^2 \implies x \, \mathrm{d}x = -t \, \mathrm{d}t$ [cite: 1361, 1362] Rewrite the internal integral block components [cite: 1363]: f(x) = int x^2 cdot sqrt3-x^2 cdot (x \, mathrmdx) = int (3-t^2) cdot t cdot (-t \, mathrmdt)$f(x) = \int x^2 \cdot \sqrt{3-x^2} \cdot (x \, \mathrm{d}x) = \int (3-t^2) \cdot t \cdot (-t \, \mathrm{d}t)$ [cite: 1363] = int (t^4 - 3t^2) \, mathrmdt = fract^55 - t^3 + C$= \int (t^4 - 3t^2) \, \mathrm{d}t = \frac{t^5}{5} - t^3 + C$ [cite: 1363, 1366] Return to original reference variable x$x$ [cite: 1366]: f(x) = frac(3-x^2)^5/25 - (3-x^2)^3/2 + C$f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} + C$ [cite: 1366] ### Step 1: Constant integration resolving Evaluate function boundary conditions at x = sqrt2$x = \sqrt{2}$ [cite: 1366]: f(sqrt2) = frac(3-2)^5/25 - (3-2)^3/2 + C = frac15 - 1 + C = -frac45 + C$f(\sqrt{2}) = \frac{(3-2)^{5/2}}{5} - (3-2)^{3/2} + C = \frac{1}{5} - 1 + C = -\frac{4}{5} + C$ [cite: 1366] Given 5f(sqrt2) = -4 implies f(sqrt2) = -frac45$5f(\sqrt{2}) = -4 \implies f(\sqrt{2}) = -\frac{4}{5}$ [cite: 638, 1366]. -frac45 + C = -frac45 implies C = 0$-\frac{4}{5} + C = -\frac{4}{5} \implies C = 0$ [cite: 1366] ### Step 2: Numeric tracking value Evaluate final targeted definition state value at x=1$x=1$ [cite: 1367]: f(1) = frac(3-1)^5/25 - (3-1)^3/2 = frac2^5/25 - 2^3/2$f(1) = \frac{(3-1)^{5/2}}{5} - (3-1)^{3/2} = \frac{2^{5/2}}{5} - 2^{3/2}$ [cite: 1367] = 2^3/2left(frac25 - 1right) = 2sqrt2left(-frac35right) = -frac6sqrt25$= 2^{3/2}\left(\frac{2}{5} - 1\right) = 2\sqrt{2}\left(-\frac{3}{5}\right) = -\frac{6\sqrt{2}}{5}$ [cite: 1367, 1368] ### Pattern Recognition Splitting powers of x$x$ to create a direct match with internal derivative differential flags speeds up the integration transformation sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals

### Related Formula Domain of log chain iterations: For log_2log_4(M) > 0$\log_2\log_4(M) > 0$, we require log_4(M) > 1 implies M > 4$\log_4(M) > 1 \implies M > 4$. ### Core Logic Trace internal arguments outward sequentially [cite: 1393, 1394]: log_4log_6(3 + 4x - x^2) > 0 implies log_6(3 + 4x - x^2) > 1$\log_{4}\log_{6}(3 + 4x - x^2) > 0 \implies \log_{6}(3 + 4x - x^2) > 1$ [cite: 1393, 1394] 3 + 4x - x^2 > 6^1 implies x^2 - 4x + 3 < 0$3 + 4x - x^2 > 6^1 \implies x^2 - 4x + 3 < 0$ [cite: 1395, 1396] (x-1)(x-3) < 0 implies x in (1, 3)$(x-1)(x-3) < 0 \implies x \in (1, 3)$ [cite: 1397, 1398] Thus, determine limits [cite: 1399]: a = 1, quad b = 3 implies b - a = 2$a = 1, \quad b = 3 \implies b - a = 2$ [cite: 1399] ### Step 1: Setting up the greatest integer function integration We need to evaluate int_0^2 [x^2] \, mathrmdx$\int_0^2 [x^2] \, \mathrm{d}x$[cite: 1400]. Identify step boundary switch locations inside range [0, 2]$[0, 2]$ [cite: 1400]: - For x in [0, 1): [x^2] = 0$x \in [0, 1): [x^2] = 0$ - For x in [1, sqrt2): [x^2] = 1$x \in [1, \sqrt{2}): [x^2] = 1$ - For x in [sqrt2, sqrt3): [x^2] = 2$x \in [\sqrt{2}, \sqrt{3}): [x^2] = 2$ - For x in [sqrt3, 2): [x^2] = 3$x \in [\sqrt{3}, 2): [x^2] = 3$ Set up separate boundary component integrations [cite: 1400]: int_0^2 [x^2] \, mathrmdx = int_0^1 0 \, mathrmdx + int_1^sqrt2 1 \, mathrmdx + int_sqrt2^sqrt3 2 \, mathrmdx + int_sqrt3^2 3 \, mathrmdx$\int_0^2 [x^2] \, \mathrm{d}x = \int_0^1 0 \, \mathrm{d}x + \int_1^{\sqrt{2}} 1 \, \mathrm{d}x + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, \mathrm{d}x + \int_{\sqrt{3}}^2 3 \, \mathrm{d}x$ [cite: 1400] = 0 + (sqrt2 - 1) + 2(sqrt3 - sqrt2) + 3(2 - sqrt3)$= 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$ [cite: 1400] = sqrt2 - 1 + 2sqrt3 - 2sqrt2 + 6 - 3sqrt3 = 5 - sqrt2 - sqrt3$= \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 5 - \sqrt{2} - \sqrt{3}$ [cite: 1400] ### Step 2: Matching coefficients Compare values with requested answer template shape [cite: 1400]: 5 - sqrt2 - sqrt3 = p - sqrtq - sqrtr$5 - \sqrt{2} - \sqrt{3} = p - \sqrt{q} - \sqrt{r}$ [cite: 1400] p = 5, quad q = 2, quad r = 3$p = 5, \quad q = 2, \quad r = 3$ [cite: 1400] textFinal Sum = p + q + r = 5 + 2 + 3 = 10$\text{Final Sum} = p + q + r = 5 + 2 + 3 = 10$ [cite: 1400] ### Pattern Recognition Integrals over greatest integer configurations change value exactly where the inner expression tracks through integer milestones. Mapping boundaries accurately resolves calculations smoothly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals

### Related Formula Definite integration geometry area: Split boundary zones around intersections where functional dominant switches occur.

### Core Logic Analyze intersection points between y_1 = |x|$y_1 = |x|$ and y_2 = x|x-2|$y_2 = x|x-2|$ across required integration span regions: - For x in [-2, 0]$x \in [-2, 0]$: |x| = -x$|x| = -x$ and x|x-2| = -x(2-x) = x^2 - 2x$x|x-2| = -x(2-x) = x^2 - 2x$. Max curve tracks through distinct segments. - For positive sectors, compute intersections: x = x(2-x) implies x = 1$x = x(2-x) \implies x = 1$ or x=0$x=0$. Also check switch locations where graphs swap dominance. ### Step 1: Setting up separate area integral blocks Using geometric area partitions calculated across continuous regions [cite: 1495]: textArea = frac12 times 2 times 2 + frac12 times 3 times 3 + frac12 times 1 times 11 = 12$\text{Area} = \frac{1}{2} \times 2 \times 2 + \frac{1}{2} \times 3 \times 3 + \frac{1}{2} \times 1 \times 11 = 12$ [cite: 1495] Alternatively, splitting the boundary metrics via continuous definite limits yields identical whole tracking blocks matching exactly to 12 total units[cite: 1495]. ### Pattern Recognition Plotting multiple curves dynamically highlights dominance shifts quickly. Computing distinct straight triangular chunks saves valuable integration time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals (Application of Integrals)

### Core Logic Let's simplify the integrand by rationalizing the denominator term block. Notice that: left(sqrt1+x^2 - xright)left(sqrt1+x^2 + xright) = (1+x^2) - x^2 = 1$\left(\sqrt{1+x^2} - x\right)\left(\sqrt{1+x^2} + x\right) = (1+x^2) - x^2 = 1$ frac1sqrt1+x^2 - x = sqrt1+x^2 + x$\frac{1}{\sqrt{1+x^2} - x} = \sqrt{1+x^2} + x$ Substituting this back into the denominator expression column: I = int left(sqrt1+x^2 + xright)^10 cdot left(sqrt1+x^2 + xright)^9 dx = int left(sqrt1+x^2 + xright)^19 dx$I = \int \left(\sqrt{1+x^2} + x\right)^{10} \cdot \left(\sqrt{1+x^2} + x\right)^9 dx = \int \left(\sqrt{1+x^2} + x\right)^{19} dx$ ### Step 1: Implementing the Substitution Path Let t = sqrt1+x^2 + x$t = \sqrt{1+x^2} + x$. Then: dt = left( fracxsqrt1+x^2 + 1 right) dx = left( fracx + sqrt1+x^2sqrt1+x^2 right) dx = fractsqrt1+x^2 dx$dt = \left( \frac{x}{\sqrt{1+x^2}} + 1 \right) dx = \left( \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}} \right) dx = \frac{t}{\sqrt{1+x^2}} dx$ dx = fracsqrt1+x^2t dt$dx = \frac{\sqrt{1+x^2}}{t} dt$ Since sqrt1+x^2 + x = t$\sqrt{1+x^2} + x = t$ and sqrt1+x^2 - x = frac1t$\sqrt{1+x^2} - x = \frac{1}{t}$, adding both gives: 2sqrt1+x^2 = t + frac1t implies sqrt1+x^2 = frac12left(t + frac1tright)$2\sqrt{1+x^2} = t + \frac{1}{t} \implies \sqrt{1+x^2} = \frac{1}{2}\left(t + \frac{1}{t}\right)$ Thus, dx = frac12tleft(t + frac1tright) dt = frac12left(1 + frac1t^2right) dt$dx = \frac{1}{2t}\left(t + \frac{1}{t}\right) dt = \frac{1}{2}\left(1 + \frac{1}{t^2}\right) dt$. ### Step 2: Integrating with respect to t Substitute these back into the integral: I = int t^19 cdot frac12left(1 + frac1t^2right) dt = frac12 int left(t^19 + t^17right) dt$I = \int t^{19} \cdot \frac{1}{2}\left(1 + \frac{1}{t^2}\right) dt = \frac{1}{2} \int \left(t^{19} + t^{17}\right) dt$ I = frac12 left( fract^2020 + fract^1818 right) + C = fract^184 left( fract^210 + frac19 right) + C = fract^18360 big(9t^2 + 10big) + C$I = \frac{1}{2} \left( \frac{t^{20}}{20} + \frac{t^{18}}{18} \right) + C = \frac{t^{18}}{4} \left( \frac{t^2}{10} + \frac{1}{9} \right) + C = \frac{t^{18}}{360} \big(9t^2 + 10\big) + C$ ### Step 3: Matching Form and Finding m + n To match the template format, let's pull out a factor of t$t$: I = fract^19360 left( 9t + frac10t right) + C = fract^19360 left( 9left(sqrt1+x^2+xright) + 10left(sqrt1+x^2-xright) right) + C$I = \frac{t^{19}}{360} \left( 9t + \frac{10}{t} \right) + C = \frac{t^{19}}{360} \left( 9\left(\sqrt{1+x^2}+x\right) + 10\left(\sqrt{1+x^2}-x\right) \right) + C$ I = fracleft(sqrt1+x^2+xright)^19360 left( 19sqrt1+x^2 - x right) + C$I = \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{360} \left( 19\sqrt{1+x^2} - x \right) + C$ Comparing this directly with the given answer format, we identify: - m = 360$m = 360$ - n = 19$n = 19$ Computing m + n$m + n$: m + n = 360 + 19 = 379$m + n = 360 + 19 = 379$ ### Pattern Recognition Expressions containing conjugate factors like sqrt1+x^2 pm x$\sqrt{1+x^2} \pm x$ frequently simplify under rationalization because their product equals 1. This dynamic quickly reduces fractional components into single power blocks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Indefinite Integrals

### Related Formula The area enclosed between upper bounding function y_textupper$y_{\text{upper}}$ and lower function y_textlower$y_{\text{lower}}$ is: textArea = int_a^b (y_textupper - y_textlower) \, dx$\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx$ ### Core Logic We need to find the intersection points of the curves to understand the min\x+7, 11-3x\$\min\{x+7, 11-3x\}$ boundary transition: 1) x+7 = 11-3x implies 4x = 4 implies x = 1$x+7 = 11-3x \implies 4x = 4 \implies x = 1$. Hence, the line switches behavior at x=1$x=1$. 2) Intersecting 1+x^2$1+x^2$ with x+7$x+7$: x^2 - x - 6 = 0 implies (x-3)(x+2) = 0 implies x = -2 quad textor quad x = 3$x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0 \implies x = -2 \quad \text{or} \quad x = 3$ 3) Intersecting 1+x^2$1+x^2$ with 11-3x$11-3x$: x^2 + 3x - 10 = 0 implies (x+5)(x-2) = 0 implies x = 2 quad textor quad x = -5$x^2 + 3x - 10 = 0 \implies (x+5)(x-2) = 0 \implies x = 2 \quad \text{or} \quad x = -5$

### Step 1: Set up Integrals The transition points show that from x = -2$x = -2$ to 1$1$, the upper line is x+7$x+7$, and from x = 1$x = 1$ to 2$2$, the upper line is 11-3x$11-3x$. A = int_-2^1 ((x + 7) - (1 + x^2)) \, dx + int_1^2 ((11 - 3x) - (1 + x^2)) \, dx$A = \int_{-2}^{1} ((x + 7) - (1 + x^2)) \, dx + \int_{1}^{2} ((11 - 3x) - (1 + x^2)) \, dx$ A = int_-2^1 (x + 6 - x^2) \, dx + int_1^2 (10 - 3x - x^2) \, dx$A = \int_{-2}^{1} (x + 6 - x^2) \, dx + \int_{1}^{2} (10 - 3x - x^2) \, dx$ ### Step 2: Integration Evaluation Evaluating the first integral: left[ fracx^22 + 6x - fracx^33 right]_-2^1 = left(frac12 + 6 - frac13right) - left(2 - 12 + frac83right) = frac376 - left(-frac223right) = frac272$\left[ \frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-2}^{1} = \left(\frac{1}{2} + 6 - \frac{1}{3}\right) - \left(2 - 12 + \frac{8}{3}\right) = \frac{37}{6} - \left(-\frac{22}{3}\right) = \frac{27}{2}$ Evaluating the second integral: left[ 10x - frac3x^22 - fracx^33 right]_1^2 = left(20 - 6 - frac83right) - left(10 - frac32 - frac13right) = frac343 - frac496 = frac196$\left[ 10x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{1}^{2} = \left(20 - 6 - \frac{8}{3}\right) - \left(10 - \frac{3}{2} - \frac{1}{3}\right) = \frac{34}{3} - \frac{49}{6} = \frac{19}{6}$ Total Area A$A$: A = frac272 + frac196 = frac81 + 196 = frac1006 = frac503$A = \frac{27}{2} + \frac{19}{6} = \frac{81 + 19}{6} = \frac{100}{6} = \frac{50}{3}$ ### Step 3: Calculate 3A Multiplying the total area by 3: 3A = 3 cdot left(frac503right) = 50$3A = 3 \cdot \left(\frac{50}{3}\right) = 50$ ### Pattern Recognition When a boundary contains a min\\$\min\{\}$ or max\\$\max\{\}$ component, always solve for their internal intersection first to identify the exact splitting point of your definite integrals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus