Integration by substitution

Use a given substitution (e.g., u = 2x + 3) to evaluate a definite integral, often showing it equals a specific exact value.

2 questions · Standard +0.8

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Edexcel C34 2017 January Q9
9 marks Standard +0.3
9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e30f0c28-1695-40a1-8e9a-6ea7e29042bf-16_727_1491_258_239} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure}
  1. By using the substitution \(u = 2 x + 3\), show that $$\int _ { 0 } ^ { 12 } \frac { x } { ( 2 x + 3 ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 2 } \ln 3 - \frac { 2 } { 9 }$$ The curve \(C\) has equation $$y = \frac { 9 \sqrt { x } } { ( 2 x + 3 ) } , \quad x > 0$$ The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\), the \(x\)-axis and the line with equation \(x = 12\). The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
  2. Use the result of part (a) to find the exact value of the volume of the solid generated.
Edexcel C4 2010 January Q8
10 marks Challenging +1.2
8. (a) Using the substitution \(x = 2 \cos u\), or otherwise, find the exact value of $$\int _ { 1 } ^ { \sqrt { 2 } } \frac { 1 } { x ^ { 2 } \sqrt { } \left( 4 - x ^ { 2 } \right) } d x$$ \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5ef3ae4a-a06d-48c1-8b79-7d7c3f95d120-14_680_1264_502_338} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = \frac { 4 } { x \left( 4 - x ^ { 2 } \right) ^ { \frac { 1 } { 4 } } } , \quad 0 < x < 2\). The shaded region \(S\), shown in Figure 3, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = 1\) and \(x = \sqrt { } 2\). The shaded region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed.