6. Use the Fundamental Theorem of Calculus to evaluate the following definite integral. \( \int_{0}^{2} 15 e^{3 \mathrm{x}} \mathrm{dx} \) \( \int_{0}^{2} 15 e^{3 \mathrm{x}} \mathrm{dx}=5 e^{6}-5 \) (Type an exact answer.)

\( \frac{d}{dx}(\ln (\arctan (x+1))-\ln (\arctan (x))) \)

a) \( \iint_{\boldsymbol{R}}(\mathbf{5}-\boldsymbol{x}-\mathbf{2 y}) \boldsymbol{d} x \boldsymbol{d y} \) b) \( \iint_{\boldsymbol{R}}(\mathbf{5}-\boldsymbol{x}-\mathbf{2 y}) \boldsymbol{d y} \boldsymbol{d x} \) \[ R=\{(x, y) \mid 1 \leq x \leq 2,0 \leq y \leq 3\} \]

container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially illed with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate f \( 0.01 \mathrm{~m}^{3} \) per second. At time \( t \) seconds the fertiliser remaining in the container forms an inverted cone of eight \( h \) metres. The volume of a cone is \( V=\frac{1}{3} \pi r^{2} h \).] (i) Show that \( h^{2} \frac{\mathrm{~d} h}{\mathrm{~d} t}=-\frac{9}{400 \pi} \). (ii) Express \( h \) in terms of \( t \).

Exercice 1: Soit \( f \) la fonction numérique définie sur \( \mathbb{R} \) par: \( f(x)=E(x)+E\left(x+\frac{1}{2}\right)-E(2 x) \) 1) Montrer que : \( \forall x \in\left[0, \frac{1}{2}[; f(x)=0\right. \). 2) Montrer que \( f \) est \( \frac{1}{2} \)-périodique. 3) Montrer que : \( \forall x \in \mathbb{R}, \exists n \in \mathbb{Z} ; x-n \frac{1}{2} \in\left[0, \frac{1}{2}[\right. \). 4) En déduire que : \( \forall x \in \mathbb{R} ; E(x)+E\left(x+\frac{1}{2}\right)=E(2 x) \).

A container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially filled with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate of \( 0.01 \mathrm{~m}^{3} \) per second. At time \( t \) seconds the fertiliser remaining in the container forms an inverted cone of height \( h \) metres. [The volume of a cone is \( V=\frac{1}{3} \pi r^{2} h \).] (i) Show that \( h^{2} \frac{\mathrm{~d} h}{\mathrm{~d} t}=-\frac{9}{400 \pi} \). (ii) Express \( h \) in terms of \( t \). (iii) Find the time it takes to empty the container, giving your answer to the nearest minute.