### Related Formula The beta function integral format satisfies: I(m,n) = int_0^1 x^m-1 (1-x)^n-1 dx$I(m,n) = \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx$ ### Core Logic Let's combine the terms of the requested sum directly by inserting their respective definitions: I(9,14) = int_0^1 x^9-1 (1-x)^14-1 dx = int_0^1 x^8 (1-x)^13 dx$I(9,14) = \int_{0}^{1} x^{9-1} (1-x)^{14-1} dx = \int_{0}^{1} x^8 (1-x)^{13} dx$ I(10,13) = int_0^1 x^10-1 (1-x)^13-1 dx = int_0^1 x^9 (1-x)^12 dx$I(10,13) = \int_{0}^{1} x^{10-1} (1-x)^{13-1} dx = \int_{0}^{1} x^9 (1-x)^{12} dx$ ### Step 1: Factoring out common algebraic terms Summing the two components: I(9,14) + I(10,13) = int_0^1 left[ x^8 (1-x)^13 + x^9 (1-x)^12 right] dx$I(9,14) + I(10,13) = \int_{0}^{1} \left[ x^8 (1-x)^{13} + x^9 (1-x)^{12} \right] dx$ Factor out the common term x^8 (1-x)^12$x^8 (1-x)^{12}$ inside the integrand: = int_0^1 x^8 (1-x)^12 left[ (1-x) + x right] dx$= \int_{0}^{1} x^8 (1-x)^{12} \left[ (1-x) + x \right] dx$ = int_0^1 x^8 (1-x)^12 (1) dx$= \int_{0}^{1} x^8 (1-x)^{12} (1) dx$ = int_0^1 x^9-1 (1-x)^13-1 dx = I(9,13)$= \int_{0}^{1} x^{9-1} (1-x)^{13-1} dx = I(9,13)$ ### Pattern Recognition When dealing with linear combinations of beta functions with shifting parameter indices, directly writing down the definite integral expression often results in immediate algebraic cancellation or simplification via basic factoring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

### 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 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