Area under curve with fractional/negative powers or roots

9 \includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-12_764_967_292_555} The diagram shows the curve with equation \(y = \sqrt { 2 x ^ { 3 } + 10 }\).

1. Find the equation of the tangent to the curve at the point where \(x = 3\). Give your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers. \includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-12_2716_35_141_2013}
2. The region shaded in the diagram is enclosed by the curve and the straight lines \(x = 1 , x = 3\) and \(y = 0\). Find the volume of the solid obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

9. \begin{figure}[h] \end{figure} Diagram not drawn to scale

1. Find $$\int \frac { 1 } { ( 2 x - 1 ) ^ { 2 } } d x$$ Figure 2 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = \frac { 12 } { ( 2 x - 1 ) } \quad 1 \leqslant x \leqslant 5$$ The finite region \(R\), shown shaded in Figure 2, is bounded by the line with equation \(x = 1\), the curve with equation \(y = \mathrm { f } ( x )\) and the line with equation \(y = \frac { 4 } { 3 }\). The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
2. Find the exact value of the volume of the solid generated, giving your answer in its simplest form.
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8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = \sqrt { x } , x \geqslant 0\) The region \(R\), shown shaded in Figure 2, is bounded by the curve, the line with equation \(x = 1\), the \(x\)-axis and the line with equation \(x = a\), where \(a\) is a constant. Given that the area of \(R\) is 10

1. find, in simplest form, the value of
   1. \(\int _ { 1 } ^ { a } \sqrt { 8 x } \mathrm {~d} x\)
   2. \(\int _ { 0 } ^ { a } \sqrt { x } \mathrm {~d} x\)
2. show that \(a = 2 ^ { k }\), where \(k\) is a rational constant to be found.

3. \begin{figure}[h] \end{figure} Figure 1 shows a circle with radius \(r\) and centre at the origin.
The region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis and the part of the circle for which \(y > 0\) The region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis to create a sphere with volume \(V\) Use integration to show that \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\)

\includegraphics{figure_5} The diagram shows the curve with equation \(y = 10x^{\frac{1}{2}} - \frac{5}{2}x^{\frac{3}{2}}\) for \(x > 0\). The curve meets the \(x\)-axis at the points \((0, 0)\) and \((4, 0)\). Find the area of the shaded region. [4]

\includegraphics{figure_7} The diagram shows part of the curve with equation \(y = \frac{12}{\sqrt{2x+1}}\). The point \(A\) on the curve has coordinates \(\left(\frac{7}{2}, 6\right)\).

1. Find the equation of the tangent to the curve at \(A\). Give your answer in the form \(y = mx + c\). [4]
2. Find the area of the region bounded by the curve and the lines \(x = 0\), \(x = \frac{7}{2}\) and \(y = 0\). [4]

\includegraphics{figure_6} The diagram shows the curve with equation \(y = \frac{1}{\sqrt{3x + 2}}\). The shaded region is bounded by the curve and the lines \(x = 0\), \(x = 2\) and \(y = 0\).

1. Find the exact area of the shaded region. [4]
2. The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed, simplifying your answer. [5]

A curve has equation \(y = 6x^2 + \frac{8}{x^2}\) and is sketched below for \(x > 0\) \includegraphics{figure_6} Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = a\) and \(x = 2a\), where \(a > 0\), giving your answer in terms of \(a\) [4 marks]

\includegraphics{figure_1} The diagram shows the curve \(y = \sqrt{x - 3}\). The shaded region is bounded by the curve and the two axes. Find the exact area of the shaded region. [4]