### Related Formula Leibniz Integral Rule for differentiation under integral sign: fracmathrmdmathrmdxleft(int_phi(x)^psi(x) f(t)mathrmdtright) = f(psi(x))psi'(x) - f(phi(x))phi'(x)$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{\phi(x)}^{\psi(x)} f(t)\mathrm{d}t\right) = f(\psi(x))\psi'(x) - f(\phi(x))\phi'(x)$ ### Core Logic Rewrite equation to isolate the integral kernel: y = 1 - 2x + e^x int_0^x e^-t f(t)mathrmdt$y = 1 - 2x + e^x \int_0^x e^{-t} f(t)\mathrm{d}t$ Differentiating with respect to x$x$ using product rule and Leibniz rule: fracmathrmdymathrmdx = -2 + e^x int_0^x e^-t f(t)mathrmdt + e^x cdot left(e^-x f(x)right)$\frac{\mathrm{d}y}{\mathrm{d}x} = -2 + e^x \int_0^x e^{-t} f(t)\mathrm{d}t + e^x \cdot \left(e^{-x} f(x)\right)$ Notice that e^x int_0^x e^-t f(t)mathrmdt = y - (1 - 2x)$e^x \int_0^x e^{-t} f(t)\mathrm{d}t = y - (1 - 2x)$. Substitute this back: fracmathrmdymathrmdx = -2 + [y - 1 + 2x] + y implies fracmathrmdymathrmdx - 2y = 2x - 3$\frac{\mathrm{d}y}{\mathrm{d}x} = -2 + [y - 1 + 2x] + y \implies \frac{\mathrm{d}y}{\mathrm{d}x} - 2y = 2x - 3$ ### Step 1: Solve Differential Equation Integrating factor textI.F. = e^int -2 mathrmdx = e^-2x$\text{I.F.} = e^{\int -2 \mathrm{d}x} = e^{-2x}$. y e^-2x = int (2x - 3)e^-2xmathrmdx = frac-(2x - 3)2e^-2x - frac12e^-2x + c$y e^{-2x} = \int (2x - 3)e^{-2x}\mathrm{d}x = \frac{-(2x - 3)}{2}e^{-2x} - \frac{1}{2}e^{-2x} + c$ From original equation, at x=0$x=0$, f(0) = 1$f(0) = 1$. Evaluating c$c$: 1 = frac32 - frac12 + c implies c = 0$1 = \frac{3}{2} - \frac{1}{2} + c \implies c = 0$ Thus, y = -x + 1 implies x + y = 1$y = -x + 1 \implies x + y = 1$. ### Step 2: Calculate Area The boundary line is x + y = 1$x + y = 1$. The area bounded by this line and coordinate axes is a right triangle with intercepts (1,0)$(1,0)$ and (0,1)$(0,1)$: textArea = frac12 times 1 times 1 = frac12$\text{Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$ ### Pattern Recognition Integral equations of convolution type (e^x-t$e^{x-t}$) always simplify directly into standard linear ordinary differential equations of first or second order when tracking Leibniz rules properly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Area Under Curves

### Related Formula For a linear differential equation fracdydx + Py = Q$\frac{dy}{dx} + Py = Q$, the Integrating Factor (IF) is defined as: textIF = e^int P \, dx$\text{IF} = e^{\int P \, dx}$ ### Core Logic Divide the full differential equation by (x^2+1)$(x^2+1)$: fracdydx - left(frac2xx^2+1right)y = frac(x^2+1)^2 cos xx^2+1 = (x^2+1)cos x$\frac{dy}{dx} - \left(\frac{2x}{x^2+1}\right)y = \frac{(x^2+1)^2 \cos x}{x^2+1} = (x^2+1)\cos x$ This is a standard Linear Differential Equation with: P = -frac2xx^2+1, quad Q = (x^2+1)cos x$P = -\frac{2x}{x^2+1}, \quad Q = (x^2+1)\cos x$ textIF = e^int -frac2xx^2+1\,dx = e^-ln(x^2+1) = frac1x^2+1$\text{IF} = e^{\int -\frac{2x}{x^2+1}\,dx} = e^{-\ln(x^2+1)} = \frac{1}{x^2+1}$ ### Step 1: Solve for General Solution The solution format is y cdot textIF = int Q cdot textIF \, dx$y \cdot \text{IF} = \int Q \cdot \text{IF} \, dx$: y cdot frac1x^2+1 = int (x^2+1)cos x cdot frac1x^2+1 \, dx$y \cdot \frac{1}{x^2+1} = \int (x^2+1)\cos x \cdot \frac{1}{x^2+1} \, dx$ fracyx^2+1 = sin x + c$\frac{y}{x^2+1} = \sin x + c$ Using the boundary condition y(0) = 1$y(0) = 1$: frac10+1 = sin(0) + c implies c = 1$\frac{1}{0+1} = \sin(0) + c \implies c = 1$ y = (x^2+1)(sin x + 1)$y = (x^2+1)(sin x + 1)$ ### Step 2: Definite Integration Evaluation We need to evaluate int_-3^3 y \, dx$\int_{-3}^{3} y \, dx$: int_-3^3 (x^2+1)(sin x + 1) \, dx = int_-3^3 (x^2sin x + x^2 + sin x + 1) \, dx$\int_{-3}^{3} (x^2+1)(sin x + 1) \, dx = \int_{-3}^{3} (x^2\sin x + x^2 + \sin x + 1) \, dx$ By symmetry of odd/even functions over symmetric intervals [-a, a]$[-a, a]$: int_-3^3 x^2sin x \, dx = 0$\int_{-3}^{3} x^2\sin x \, dx = 0$ (since it is an odd function) int_-3^3 sin x \, dx = 0$\int_{-3}^{3} \sin x \, dx = 0$ (since it is an odd function) Thus, we are left with the even components: int_-3^3 (x^2 + 1) \, dx = 2 int_0^3 (x^2 + 1) \, dx = 2 left[ fracx^33 + x right]_0^3 = 2(9 + 3) = 24$\int_{-3}^{3} (x^2 + 1) \, dx = 2 \int_{0}^{3} (x^2 + 1) \, dx = 2 \left[ \frac{x^3}{3} + x \right]_{0}^{3} = 2(9 + 3) = 24$ ### Pattern Recognition Splitting a symmetric interval integral into odd and even parts immediately simplifies calculations by dropping all odd functions down to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integral Calculus

### Related Formula Standard first-order linear differential equation structure: fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$ textIntegrating Factor (I.F.) = e^int P(x)dx$\text{Integrating Factor (I.F.)} = e^{\int P(x)dx}$ ### Step 1: Reduce into Standard Format Divide the full expression by 2cos x$2\cos x$ [cite: 3411, 4085]: fracdydx = frac2sin x cos x2cos x - frac4ysin x2cos x$\frac{dy}{dx} = \frac{2\sin x \cos x}{2\cos x} - \frac{4y\sin x}{2\cos x}$ fracdydx + 2ytan x = sin x$\frac{dy}{dx} + 2y\tan x = \sin x$ ### Step 2: Integrating Factor & Solution Compute the integrating multiplier : textI.F. = e^int 2tan x dx = e^2ln|sec x| = sec^2 x$\text{I.F.} = e^{\int 2\tan x dx} = e^{2\ln|\sec x|} = \sec^2 x$ Write general integration solution path : y cdot sec^2 x = int sin x cdot sec^2 x dx = int sec x tan x dx = sec x + C$y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx = \int \sec x \tan x dx = \sec x + C$ y = cos x + Ccos^2 x$y = \cos x + C\cos^2 x$ ### Step 3: Boundary Evaluation Apply the initialization condition yleft(fracpi3right) = 0$y\left(\frac{\pi}{3}\right) = 0$ [cite: 3412, 4087]: 0 = cosleft(fracpi3right) + Ccos^2left(fracpi3right) Rightarrow 0 = frac12 + Cleft(frac14right) Rightarrow C = -2$0 = \cos\left(\frac{\pi}{3}\right) + C\cos^2\left(\frac{\pi}{3}\right) \Rightarrow 0 = \frac{1}{2} + C\left(\frac{1}{4}\right) \Rightarrow C = -2$ Thus, the solution is y = cos x - 2cos^2 x$y = \cos x - 2\cos^2 x$ . Find derivative yprime$y\prime$ : yprime = -sin x + 4cos x sin x = -sin x + 2sin 2x$y\prime = -\sin x + 4\cos x \sin x = -\sin x + 2\sin 2x$ ### Step 4: Target Calculation Evaluate components at x = fracpi4$x = \frac{\pi}{4}$ : yleft(fracpi4right) = frac1sqrt2 - 2left(frac12right) = frac1sqrt2 - 1$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - 2\left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}} - 1$ yprimeleft(fracpi4right) = -frac1sqrt2 + 2sinleft(fracpi2right) = -frac1sqrt2 + 2$y\prime\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} + 2\sin\left(\frac{\pi}{2}\right) = -\frac{1}{\sqrt{2}} + 2$ yprimeleft(fracpi4right) + yleft(fracpi4right) = left(-frac1sqrt2 + 2right) + left(frac1sqrt2 - 1right) = 1$y\prime\left(\frac{\pi}{4}\right) + y\left(\frac{\pi}{4}\right) = \left(-\frac{1}{\sqrt{2}} + 2\right) + \left(\frac{1}{\sqrt{2}} - 1\right) = 1$ ### Pattern Recognition Linear standard layout conversions depend entirely on clear integrating factor reductions. Remember int tan x dx = ln|sec x|$\int \tan x dx = \ln|\sec x|$ clearly to safely output exact matching polynomial definitions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula A linear differential equation of the first order matching fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$ uses an Integrating Factor written as: textI.F. = e^int P(x) dx$\text{I.F.} = e^{\int P(x) dx}$ ### Core Logic Rearrange the given differential equation terms to express it in standard linear form: (1 + x^2) fracdydx + xy = 5x^2sqrt1 + x^2$(1 + x^2) \frac{dy}{dx} + xy = 5x^2\sqrt{1 + x^2}$ fracdydx + left(fracx1+x^2right)y = frac5x^2sqrt1+x^2$\frac{dy}{dx} + \left(\frac{x}{1+x^2}\right)y = \frac{5x^2}{\sqrt{1+x^2}}$ ### Step 1: Compute Integrating Factor Calculate the exponent integral for textI.F.$\text{I.F.}$: int P(x) dx = int fracx1+x^2 dx = frac12 ln(1+x^2) = lnsqrt1+x^2$\int P(x) dx = \int \frac{x}{1+x^2} dx = \frac{1}{2} \ln(1+x^2) = \ln\sqrt{1+x^2}$ textI.F. = e^lnsqrt1+x^2 = sqrt1+x^2$\text{I.F.} = e^{\ln\sqrt{1+x^2}} = \sqrt{1+x^2}$ ### Step 2: General Solution and Boundary Condition The general solution template is: y cdot (textI.F.) = int Q(x) cdot (textI.F.) dx$y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx$ ysqrt1+x^2 = int frac5x^2sqrt1+x^2 cdot sqrt1+x^2 dx$y\sqrt{1+x^2} = \int \frac{5x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} dx$ ysqrt1+x^2 = int 5x^2 dx = frac5x^33 + C$y\sqrt{1+x^2} = \int 5x^2 dx = \frac{5x^3}{3} + C$ Apply the initial condition y(0) = 0$y(0) = 0$: 0 cdot sqrt1+0 = 0 + C implies C = 0$0 \cdot \sqrt{1+0} = 0 + C \implies C = 0$ Thus, the explicit functional equation is: y = frac5x^33sqrt1+x^2$y = \frac{5x^3}{3\sqrt{1+x^2}}$ ### Step 3: Evaluate at Target Value Substitute x = sqrt3$x = \sqrt{3}$ into the isolated function: y(sqrt3) = frac5(sqrt3)^33sqrt1+(sqrt3)^2 = frac5(3sqrt3)3sqrt1+3 = frac15sqrt33 cdot 2 = frac5sqrt32$y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+(\sqrt{3})^2}} = \frac{5(3\sqrt{3})}{3\sqrt{1+3}} = \frac{15\sqrt{3}}{3 \cdot 2} = \frac{5\sqrt{3}}{2}$ ### Pattern Recognition Spotting that multiplying across the differential equation format by sqrt1+x^2$\sqrt{1+x^2}$ converts the left hand side into a direct product rule derivative matching fracddx(ysqrt1+x^2)$\frac{d}{dx}(y\sqrt{1+x^2})$ yields a direct integration pathway. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations