A Calculus proof shows that \( 1=2 \). Here is the argument. By definition, \( \begin{array}{l}2+2=2^{2} \\ 3+3+3=3^{2} \\ 4+4+4+4=4^{2} \text {, and so on. } \\ \text { In general, } x+x+\ldots+x=x^{2} \text {, with } x \\ \text { number of } x^{\prime} \text { s on the left side of the } \\ \text { equation. Differentiating both sides of } \\ \text { the equation, } 1+1+\ldots+1=2 x \text {. } \\ \text { Since there are } x \text { number of } 1 \text { 's on the } \\ \text { left side of the equation, } x=2 x \text {. } \\ \text { Dividing by } x \text { on each side, } 1=2 \text {. } \\ \text { Which is the erroneous step? Explain. }\end{array} \).

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

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?

(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

\( \int _{0}^{\frac{\pi }{2}}\frac{1}{2+\cos (x)} d x \)