Use the substitution \( u=x^{2}+7 \) to find the following indefinite integral. Check your answer by differentiation. \[ \int 2 x\left(x^{2}+7\right)^{4} d x \] \( \int 2 x\left(x^{2}+7\right)^{4} d x=\frac{\left(x^{2}+7\right)^{5}}{5}+C \)

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.

What is the arc length of the parametrically defined curve given by \( x(t) = t, y(t) = t^3 \) for \( t \) in the interval \([0, 2]\)?

\( y ^ { \prime \prime } + 5 y ^ { \prime } + 4 y = 3 + 8 x ^ { 2 } + 2 \cos ( 2 x ) \)

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 \( t \).

\( \int _ { 5 + 5 - 10 } ^ { \infty } \frac { 10 \sin ( x ) } { 2 x } d x = \)