If I(m,n) = int_0^1 x^m-1 (1-x)^n-1 dx$I(m,n) = \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx$ where m, n > 0$m, n > 0$, then I(9,14) + I(10,13)$I(9,14) + I(10,13)$ is :

### Related Formula The Leibniz Integral Rule template allows direct differentiation of an integral with variable limits: fracddxleft( int_0^x g(t) dt right) = g(x)$$\frac{d}{dx}\left( \int_{0}^{x} g(t) dt \right) = g(x)$$ ### Core Logic Differentiate both sides of the given functional equation with respect to x$x$ using the product rule: fracddxleft[ 2(x+2)^2 f(x) - 3(x+2)^2 right] = fracddxleft[ 10int_0^x(t+2)f(t)dt right]$$\frac{d}{dx}\left[ 2(x+2)^2 f(x) - 3(x+2)^2 \right] = \frac{d}{dx}\left[ 10\int_{0}^{x}(t+2)f(t)dt \right]$$ 4(x+2)f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2)f(x)$$4(x+2)f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2)f(x)$$ Since x geq 0$x \geq 0$, the factor (x+2)$(x+2)$ is strictly non-zero. Divide the entire equation by 2(x+2)$2(x+2)$: 2f(x) + (x+2)f'(x) - 3 = 5f(x)$$2f(x) + (x+2)f'(x) - 3 = 5f(x)$$ (x+2)f'(x) - 3f(x) = 3$$(x+2)f'(x) - 3f(x) = 3$$ ### Step 1: Solve the First-Order Differential Equation Rearrange the expression into standard linear differential equation form where y = f(x)$y = f(x)$: fracdydx - frac3x+2y = frac3x+2$$\frac{dy}{dx} - \frac{3}{x+2}y = \frac{3}{x+2}$$ Compute the Integrating Factor (I.F.): textI.F. = e^int -frac3x+2 dx = e^-3ln(x+2) = (x+2)^-3$$\text{I.F.} = e^{\int -\frac{3}{x+2} dx} = e^{-3\ln(x+2)} = (x+2)^{-3}$$ Multiply through by the I.F. and integrate: y cdot (x+2)^-3 = int frac3x+2 cdot (x+2)^-3 dx = int 3(x+2)^-4 dx$$y \cdot (x+2)^{-3} = \int \frac{3}{x+2} \cdot (x+2)^{-3} dx = \int 3(x+2)^{-4} dx$$ fracf(x)(x+2)^3 = 3 cdot frac(x+2)^-3-3 + C = -(x+2)^-3 + C$$\frac{f(x)}{(x+2)^3} = 3 \cdot \frac{(x+2)^{-3}}{-3} + C = -(x+2)^{-3} + C$$ f(x) = -1 + C(x+2)^3$$f(x) = -1 + C(x+2)^3$$ ### Step 2: Apply the Boundary Condition Find the boundary condition by substituting x = 0$x = 0$ into the original integral equation equation: 2(0+2)^2 f(0) - 3(0+2)^2 = 10 int_0^0 (t+2)f(t) dt$$2(0+2)^2 f(0) - 3(0+2)^2 = 10 \int_{0}^{0} (t+2)f(t) dt$$ 8f(0) - 12 = 0 implies f(0) = frac128 = frac32$$8f(0) - 12 = 0 \implies f(0) = \frac{12}{8} = \frac{3}{2}$$ Substitute x = 0$x = 0$ into our general solution formula: f(0) = -1 + C(0+2)^3 implies frac32 = -1 + 8C$$f(0) = -1 + C(0+2)^3 \implies \frac{3}{2} = -1 + 8C$$ frac52 = 8C implies C = frac516$$\frac{5}{2} = 8C \implies C = \frac{5}{16}$$ Thus, the explicit function is: f(x) = -1 + frac516(x+2)^3$$f(x) = -1 + \frac{5}{16}(x+2)^3$$ ### Step 3: Evaluate at target point x = 2 Substitute x = 2$x = 2$ into the final function equation: f(2) = -1 + frac516(2+2)^3 = -1 + frac516(64)$$f(2) = -1 + \frac{5}{16}(2+2)^3 = -1 + \frac{5}{16}(64)$$ f(2) = -1 + 5(4) = -1 + 20 = 19$$f(2) = -1 + 5(4) = -1 + 20 = 19$$ ### Pattern Recognition When an equation contains a variable integral limit int_0^x$\int_0^x$, differentiating both sides using the Leibniz rule converts it into a standard differential equation. The initial value is found by setting x = 0$x = 0$ directly in the original expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

### Related Formula Integration by Parts formula: int u cdot v \, dx = u int v \, dx - int left( u' int v \, dx right) dx$$\int u \cdot v \, dx = u \int v \, dx - \int \left( u' \int v \, dx \right) dx$$ ### Core Logic Given F(x) = xf(x)$F(x) = xf(x)$ and f(2) = 1 implies F(2) = 2f(2) = 2$f(2) = 1 \implies F(2) = 2f(2) = 2$. Let's apply Integration by Parts to the first given integral: int_0^2 x F'(x) dx = 6$$\int_{0}^{2} x F'(x) dx = 6$$ left[ xF(x) right]_0^2 - int_0^2 F(x) dx = 6$$\left[ xF(x) \right]_0^2 - \int_{0}^{2} F(x) dx = 6$$ 2F(2) - 0 - int_0^2 F(x) dx = 6$$2F(2) - 0 - \int_{0}^{2} F(x) dx = 6$$ 2(2) - int_0^2 F(x) dx = 6 implies int_0^2 F(x) dx = 4 - 6 = -2$$2(2) - \int_{0}^{2} F(x) dx = 6 \implies \int_{0}^{2} F(x) dx = 4 - 6 = -2$$ ### Step 1: Simplify Second Integral via Integration by Parts Now look at the second integral: int_0^2 x^2 F''(x) dx = 40$$\int_{0}^{2} x^2 F''(x) dx = 40$$ Applying Integration by parts (taking u = x^2$u = x^2$ and v = F''(x)$v = F''(x)$): left[ x^2 F'(x) right]_0^2 - int_0^2 2x F'(x) dx = 40$$\left[ x^2 F'(x) \right]_0^2 - \int_{0}^{2} 2x F'(x) dx = 40$$ 4 F'(2) - 0 - 2 int_0^2 x F'(x) dx = 40$$4 F'(2) - 0 - 2 \int_{0}^{2} x F'(x) dx = 40$$ We already know int_0^2 x F'(x) dx = 6$\int_{0}^{2} x F'(x) dx = 6$: 4 F'(2) - 2(6) = 40$$4 F'(2) - 2(6) = 40$$ 4 F'(2) - 12 = 40 implies 4 F'(2) = 52 implies F'(2) = 13$$4 F'(2) - 12 = 40 \implies 4 F'(2) = 52 \implies F'(2) = 13$$ ### Step 2: Sum the Values We need to find F'(2) + int_0^2 F(x) dx$F'(2) + \int_{0}^{2} F(x) dx$: 13 + (-2) = 11$13 + (-2) = 11$ ### Pattern Recognition Notice how the definition of f(x)$f(x)$ is mostly a distraction to find F(2)=2$F(2)=2$. The problem is fundamentally testing consecutive applications of integration by parts to reduction structures. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration

### Related Formula int fracdta^2 - b^2 t^2 = frac12a ln left| fraca+bta-bt right|$$\int \frac{dt}{a^2 - b^2 t^2} = \frac{1}{2a} \ln \left| \frac{a+bt}{a-bt} \right|$$ sin 2theta = 1 - (sintheta - costheta)^2$$\sin 2\theta = 1 - (\sin\theta - \cos\theta)^2$$ ### Core Logic Let sintheta - costheta = t$\sin\theta - \cos\theta = t$. Then (costheta + sintheta)dtheta = dt$(\cos\theta + \sin\theta)d\theta = dt$. Transform limits: When theta = 0 implies t = 0 - 1 = -1$\theta = 0 \implies t = 0 - 1 = -1$ When theta = fracpi4 implies t = frac1sqrt2 - frac1sqrt2 = 0$\theta = \frac{\pi}{4} \implies t = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$ ### Step 1: Perform the algebraic substitution Express the denominator base: 9 + 16sin 2theta = 9 + 16[1 - t^2] = 25 - 16t^2$$9 + 16\sin 2\theta = 9 + 16[1 - t^2] = 25 - 16t^2$$ The integral transforms to: I = 80 int_-1^0 fracdt25 - 16t^2 = frac8016 int_-1^0 fracdtleft(frac54right)^2 - t^2$$I = 80 \int_{-1}^{0} \frac{dt}{25 - 16t^2} = \frac{80}{16} \int_{-1}^{0} \frac{dt}{\left(\frac{5}{4}\right)^2 - t^2}$$ ### Step 2: Execute Integral Calculation I = 5 left[ frac12left(frac54right) ln left| fracfrac54 + tfrac54 - t right| right]_-1^0$$I = 5 \left[ \frac{1}{2\left(\frac{5}{4}\right)} \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^{0}$$ I = 2 left[ ln(1) - lnleft( frac1/49/4 right) right] = 2 left[ 0 - lnleft(frac19right) right] = 2ln(9) = 4ln(3)$$I = 2 \left[ \ln(1) - \ln\left( \frac{1/4}{9/4} \right) \right] = 2 \left[ 0 - \ln\left(\frac{1}{9}\right) \right] = 2\ln(9) = 4\ln(3)$$ ### Pattern Recognition Whenever (sintheta + costheta)$(\sin\theta + \cos\theta)$ sits inside the numerator, instantly choose t = sintheta - costheta$t = \sin\theta - \cos\theta$ as your core linear substitution engine to clean denominators. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

### Related Formula textLeibniz Integral Rule for differentiation: fracddxint_0^x f(t) \, dt = f(x)$$\text{Leibniz Integral Rule for differentiation: } \frac{d}{dx}\int_{0}^{x} f(t) \, dt = f(x)$$ ### Core Logic Perform variable substitution inside the integral: let lambda x = t implies dlambda = frac1x dt$\lambda x = t \implies d\lambda = \frac{1}{x} dt$. When lambda = 0 implies t = 0$\lambda = 0 \implies t = 0$; when lambda = 1 implies t = x$\lambda = 1 \implies t = x$. The equation transforms to: frac1x int_0^x f(t) \, dt = a f(x) implies int_0^x f(t) \, dt = a x f(x)$$\frac{1}{x} \int_{0}^{x} f(t) \, dt = a f(x) \implies \int_{0}^{x} f(t) \, dt = a x f(x)$$ ### Step 1: Differentiate with respect to x Using Leibniz rule and product rule: f(x) = a [x f'(x) + f(x)]$$f(x) = a [x f'(x) + f(x)]$$ (1 - a)f(x) = a x f'(x) implies fracf'(x)f(x) = frac1-aa frac1x$$(1 - a)f(x) = a x f'(x) \implies \frac{f'(x)}{f(x)} = \frac{1-a}{a} \frac{1}{x}$$ Integrating both sides yields: ln f(x) = left(frac1-aaright)ln x + c implies f(x) = C x^frac1-aa$$\ln f(x) = \left(\frac{1-a}{a}\right)\ln x + c \implies f(x) = C x^{\frac{1-a}{a}}$$ ### Step 2: Calculate Constants using boundaries Given f(1) = 1 implies C = 1$f(1) = 1 \implies C = 1$. Given f(16) = frac18 implies frac18 = (16)^frac1-aa implies 2^-3 = (2^4)^frac1-aa$f(16) = \frac{1}{8} \implies \frac{1}{8} = (16)^{\frac{1-a}{a}} \implies 2^{-3} = (2^4)^{\frac{1-a}{a}}$ -3 = frac4(1-a)a implies -3a = 4 - 4a implies a = 4$$-3 = \frac{4(1-a)}{a} \implies -3a = 4 - 4a \implies a = 4$$ Therefore, power exponent = frac1-44 = -frac34 implies f(x) = x^-frac34$= \frac{1-4}{4} = -\frac{3}{4} \implies f(x) = x^{-\frac{3}{4}}$. ### Step 3: Evaluate target derivative value Find the derivative: f'(x) = -frac34 x^-frac74$$f'(x) = -\frac{3}{4} x^{-\frac{7}{4}}$$ Substitute x = frac116$x = \frac{1}{16}$: f'left(frac116right) = -frac34 left(2^-4right)^-frac74 = -frac34 left(2^7right) = -frac34 times 128 = -96$$f'\left(\frac{1}{16}\right) = -\frac{3}{4} \left(2^{-4}\right)^{-\frac{7}{4}} = -\frac{3}{4} \left(2^7\right) = -\frac{3}{4} \times 128 = -96$$ Final requested computation calculation: 16 - f'left(frac116right) = 16 - (-96) = 112$$16 - f'\left(\frac{1}{16}\right) = 16 - (-96) = 112$$ ### Pattern Recognition Scaling inputs inside functional definite integrals tracks closely to homogenous Euler equation properties. Converting integrations quickly to local power functions reduces processing parameters. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

### Related Formula King's Property of Definite Integrals: int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx$$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$$ ### Core Logic Let the given integral be I$I$: I = int_0^fracpi4fracx \, dxsin^4(2x)+cos^4(2x)$$I = \int_{0}^{\frac{\pi}{4}}\frac{x \, dx}{\sin^{4}(2x)+\cos^{4}(2x)}$$ Substitute 2x = t implies 2dx = dt implies dx = frac12dt$2x = t \implies 2dx = dt \implies dx = \frac{1}{2}dt$. When x = 0 implies t = 0$x = 0 \implies t = 0$, and when x = fracpi4 implies t = fracpi2$x = \frac{\pi}{4} \implies t = \frac{\pi}{2}$. I = frac14 int_0^fracpi2 fract \, dtsin^4t + cos^4t quad implies (1) $$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t \, dt}{\sin^{4}t + \cos^{4}t} \quad \implies (1) $$ ### Step 1: Apply King's Property Applying the property int_0^a f(t) \, dt = int_0^a f(a-t) \, dt$\int_{0}^{a} f(t) \, dt = \int_{0}^{a} f(a-t) \, dt$: I = frac14 int_0^fracpi2 fracleft(fracpi2 - tright) dtsin^4left(fracpi2-tright) + cos^4left(fracpi2-tright)$$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - t\right) dt}{\sin^{4}\left(\frac{\pi}{2}-t\right) + \cos^{4}\left(\frac{\pi}{2}-t\right)}$$ I = frac14 int_0^fracpi2 fracleft(fracpi2 - tright) dtcos^4t + sin^4t quad implies (2) $$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - t\right) dt}{\cos^{4}t + \sin^{4}t} \quad \implies (2) $$ Adding equations (1) and (2): 2I = frac14 int_0^fracpi2 fracfracpi2 dtsin^4t + cos^4t$$2I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} dt}{\sin^{4}t + \cos^{4}t}$$ 2I = fracpi8 int_0^fracpi2 fracdtsin^4t + cos^4t $$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sin^{4}t + \cos^{4}t} $$ 2I = fracpi8 int_0^fracpi2 fracsec^4t \, dttan^4t + 1 $$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{\sec^{4}t \, dt}{\tan^{4}t + 1} $$ 2I = fracpi8 int_0^fracpi2 frac(1 + tan^2t)sec^2t \, dttan^4t + 1$$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{(1 + \tan^{2}t)\sec^{2}t \, dt}{\tan^{4}t + 1}$$ ### Step 2: Substitution and Algebraic Limits Let tan t = y implies sec^2t \, dt = dy$\tan t = y \implies \sec^{2}t \, dt = dy$. When t = 0 implies y = 0$t = 0 \implies y = 0$, and when t = fracpi2 implies y = infty$t = \frac{\pi}{2} \implies y = \infty$. 2I = fracpi8 int_0^infty frac(1 + y^2) \, dy1 + y^4 $$2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{(1 + y^{2}) \, dy}{1 + y^{4}} $$ I = fracpi16 int_0^infty frac1 + frac1y^2y^2 + frac1y^2 \, dy $$I = \frac{\pi}{16} \int_{0}^{\infty} \frac{1 + \frac{1}{y^{2}}}{y^{2} + \frac{1}{y^{2}}} \, dy $$ Now put y - frac1y = p implies left(1 + frac1y^2right) dy = dp$y - \frac{1}{y} = p \implies \left(1 + \frac{1}{y^{2}}\right) dy = dp$. When y to 0^+ implies p to -infty$y \to 0^+ \implies p \to -\infty$, and when y to infty implies p to infty$y \to \infty \implies p \to \infty$. Also, y^2 + frac1y^2 = p^2 + 2 = p^2 + (sqrt2)^2$y^{2} + \frac{1}{y^{2}} = p^{2} + 2 = p^{2} + (\sqrt{2})^{2}$. I = fracpi16 int_-infty^infty fracdpp^2 + (sqrt2)^2 $$I = \frac{\pi}{16} \int_{-\infty}^{\infty} \frac{dp}{p^{2} + (\sqrt{2})^{2}} $$ I = fracpi16sqrt2 left[ tan^-1left(fracpsqrt2right) right]_-infty^infty $$I = \frac{\pi}{16\sqrt{2}} \left[ \tan^{-1}\left(\frac{p}{\sqrt{2}}\right) \right]_{-\infty}^{\infty} $$ $I = fracpi16sqrt2 left( fracpi2 - left(-fracpi2right) right) = fracpi^216sqrt2 = fracsqrt2pi^232 $I = \frac{\pi}{16\sqrt{2}} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \frac{\pi^{2}}{16\sqrt{2}} = \frac{\sqrt{2}\pi^{2}}{32} $ ### Pattern Recognition Sees: Integrand containing x$x$ in the numerator and symmetric trigonometric functions in the denominator. Shortcut: The elimination of x$x$ using King's property is standard. For integrals containing frac1+y^21+y^4$\frac{1+y^2}{1+y^4}$, dividing by y^2$y^2$ transforms the denominator into a perfect square form left(y-frac1yright)^2+2$\left(y-\frac{1}{y}\right)^2+2$, making substitution trivial. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 11 Mathematics: Trigonometric Identities