Chứng minh rằng \( \frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\ldots+\frac{4047}{2023^{2} \cdot 2024^{2}}<1 \)

Application 3: Simplifier les expressions suivantes: \( \begin{array}{ll}A=\ln (\sqrt{3})+\ln 3-\ln 9 & ; \\ C=\ln \left(\frac{1}{3}\right)+\ln \left(\frac{3}{5}\right)+\ln \left(\frac{5}{7}\right)+\ln \left(\frac{7}{9}\right) & ;\end{array} \quad D=\ln (\sqrt[3]{2}-\ln 16 \)

(iv) In integral equation of the form \( \int_{0}^{1} \frac{F(u) d u}{(t-u)^{n}}=G(t) \) is called. (a) Exact equation (b) Convolution type (c) Integro-differential equation (d) Abel's equation

1) Sout \( x>0 \). En appliquant le thécorème des accroissement fines à la fonction \( \ln \) montrer que: \( \exists c \in] x, x+1\left[\frac{1}{c}=\ln \left(1+\frac{1}{x}\right)\right. \) 2) \( M q: \exists c \in] 0, x\left[\right. \) arctan \( (x)=\frac{x}{1+c^{2}} \) 3) En déduire que \( \forall x>0: \) \( \frac{1}{1+x}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x} \)

A solid is formed with triangular cross-sections where each triangle's base is defined by the equation \(y = x^2\) on interval \([0, 2]\). What is the volume of this solid?

Calcula el área entre las curvas \( y = x^3 - 3x \) y \( y = x + 1 \) sobre el intervalo definido por sus intersecciones.