1.08f Area between two curves: using integration

16. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve with equation \(y = 2 x ^ { 2 } - 11 x + 12\). The curve crosses the \(y\)-axis at the point \(A\) and crosses the \(x\)-axis at the points \(B\) and \(C\).

1. Find the coordinates of the points \(A , B\) and \(C\). The point \(D\) lies on the curve such that the line \(A D\) is parallel to the \(x\)-axis. The finite region \(R\), shown shaded in Figure 4, is bounded by the curve, the line \(A C\) and the line \(A D\).
2. Use algebraic integration to find the exact area of \(R\).

10. \begin{figure}[h] \end{figure} The finite region \(R\), which is shown shaded in Figure 1, is bounded by the coordinate axes, the straight line \(l\) with equation \(y = \frac { 1 } { 3 } x + 5\) and the curve \(C\) with equation \(y = 4 x ^ { \frac { 1 } { 2 } } - x + 5 , x \geqslant 0\) The line \(l\) meets the curve \(C\) at the point \(D\) on the \(y\)-axis and at the point \(E\), as shown in Figure 1.

1. Use algebra to find the coordinates of the points \(D\) and \(E\). The curve \(C\) crosses the \(x\)-axis at the point \(F\).
2. Verify that the \(x\) coordinate of \(F\) is 25
3. Use algebraic integration to find the exact area of the shaded region \(R\).

7. The curve \(C\) has equation \(y = \mathrm { f } ( x ) , x > 0\), where $$\mathrm { f } ^ { \prime } ( x ) = 30 + \frac { 6 - 5 x ^ { 2 } } { \sqrt { x } }$$ Given that the point \(P ( 4 , - 8 )\) lies on \(C\),

1. find the equation of the tangent to \(C\) at \(P\), giving your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants.
2. Find \(\mathrm { f } ( x )\), giving each term in its simplest form.

9. \begin{figure}[h] \end{figure} Figure 2 shows

• the curve \(C\) with equation \(y = x - x ^ { 2 }\)
• the line \(l\) with equation \(y = m x\), where \(m\) is a constant and \(0 < m < 1\)

The line and the curve intersect at the origin \(O\) and at the point \(P\).

1. Find, in terms of \(m\), the coordinates of \(P\). The region \(R _ { 1 }\), shown shaded in Figure 2, is bounded by \(C\) and \(l\).
2. Show that the area of \(R _ { 1 }\) is $$\frac { ( 1 - m ) ^ { 3 } } { 6 }$$ The region \(R _ { 2 }\), also shown shaded in Figure 2, is bounded by \(C\), the \(x\)-axis and \(l\). Given that the area of \(R _ { 1 }\) is equal to the area of \(R _ { 2 }\)
3. find the exact value of \(m\). \includegraphics[max width=\textwidth, alt={}, center]{59c9f675-e7eb-47b9-b233-dfbe1844f792-33_108_76_2613_1875} \includegraphics[max width=\textwidth, alt={}, center]{59c9f675-e7eb-47b9-b233-dfbe1844f792-33_52_83_2722_1850}

1. In this question you must show all stages of your working.

\section*{Solutions based entirely on calculator technology are not acceptable.} \begin{figure}[h] \end{figure} Figure 3 shows

• the curve \(C\) with equation \(y = x ^ { 2 } - 4 x + 5\)
• the line \(l\) with equation \(y = 2\)

The curve \(C\) intersects the \(y\)-axis at the point \(D\).

1. Write down the coordinates of \(D\). The curve \(C\) intersects the line \(l\) at the points \(E\) and \(F\), as shown in Figure 3.
2. Find the \(x\) coordinate of \(E\) and the \(x\) coordinate of \(F\). Shown shaded in Figure 3 is
   • the region \(R _ { 1 }\) which is bounded by \(C , l\) and the \(y\)-axis
   • the region \(R _ { 2 }\) which is bounded by \(C\) and the line segments \(E F\) and \(D F\)
   Given that \(\frac { \text { area of } R _ { 1 } } { \text { area of } R _ { 2 } } = k\), where \(k\) is a constant,
3. use algebraic integration to find the exact value of \(k\), giving your answer as a simplified fraction.

1. The curve \(C\) has equation

$$y = \frac { ( x - k ) ^ { 2 } } { \sqrt { x } } \quad x > 0$$ where \(k\) is a positive constant.

1. Show that $$\int _ { 1 } ^ { 16 } \frac { ( x - k ) ^ { 2 } } { \sqrt { x } } \mathrm {~d} x = a k ^ { 2 } + b k + \frac { 2046 } { 5 }$$ where \(a\) and \(b\) are integers to be found. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0e3b364c-151b-471d-acb6-01afb018fb75-26_645_670_904_699} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows a sketch of the curve \(C\) and the line \(l\).
   Given that \(l\) intersects \(C\) at the point \(A ( 1,9 )\) and at the point \(B ( 16 , q )\) where \(q\) is a constant,
2. show that \(k = 4\) The region \(R\), shown shaded in Figure 1, is bounded by \(C\) and \(l\) Using the answers to parts (a) and (b),
3. find the area of region \(R\)

9. \begin{figure}[h] \end{figure} Figure 1 is a sketch of the curve \(C\) with equation $$y = 2 x ^ { \frac { 3 } { 2 } } ( 4 - x ) \quad x \geqslant 0$$ The point \(P\) is the stationary point of \(C\).

1. Find, using calculus, the \(x\) coordinate of \(P\). The region \(R _ { 1 }\), shown shaded in Figure 1, is bounded by \(C\) and the \(x\)-axis.
   The region \(R _ { 2 }\), also shown shaded in Figure 1, is bounded by \(C\), the \(x\)-axis and the line with equation \(x = k\), where \(k\) is a constant. Given that the area of \(R _ { 1 }\) is equal to the area of \(R _ { 2 }\)
2. find, using calculus, the exact value of \(k\).

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curves \(C _ { 1 }\) and \(C _ { 2 }\) with equations $$\begin{array} { l l } C _ { 1 } : y = x ^ { 3 } - 6 x + 9 & x \geqslant 0 \\ C _ { 2 } : y = - 2 x ^ { 2 } + 7 x - 1 & x \geqslant 0 \end{array}$$ The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the points \(A\) and \(B\) as shown in Figure 1 .
The point \(A\) has coordinates (1,4). Using algebra and showing all steps of your working,

1. find the coordinates of the point \(B\). The finite region \(R\), shown shaded in Figure 1, is bounded by \(C _ { 1 }\) and \(C _ { 2 }\)
2. Use algebraic integration to find the exact area of \(R\).

9. \begin{figure}[h] \end{figure} Figure 3 shows the shaded region \(R\) which is bounded by the curve \(y = - 2 x ^ { 2 } + 4 x\) and the line \(y = \frac { 3 } { 2 }\). The points \(A\) and \(B\) are the points of intersection of the line and the curve. Find

1. the \(x\)-coordinates of the points \(A\) and \(B\),
2. the exact area of \(R\).

7. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = x ( x - 1 ) ( x - 5 )$$ Use calculus to find the total area of the finite region, shown shaded in Figure 1, that is between \(x = 0\) and \(x = 2\) and is bounded by \(C\), the \(x\)-axis and the line \(x = 2\).
(9)

9. \begin{figure}[h] \end{figure} The straight line with equation \(y = x + 4\) cuts the curve with equation \(y = - x ^ { 2 } + 2 x + 24\) at the points \(A\) and \(B\), as shown in Figure 3.

1. Use algebra to find the coordinates of the points \(A\) and \(B\). The finite region \(R\) is bounded by the straight line and the curve and is shown shaded in Figure 3.
2. Use calculus to find the exact area of \(R\).

5. \begin{figure}[h] \end{figure} Figure 2 shows the line with equation \(y = 10 - x\) and the curve with equation \(y = 10 x - x ^ { 2 } - 8\) The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). The shaded area \(R\) is bounded by the line and the curve, as shown in Figure 2.
2. Calculate the exact area of \(R\).

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 1 } { 8 } x ^ { 3 } + \frac { 3 } { 4 } x ^ { 2 } , \quad x \in \mathbb { R }$$ The curve \(C\) has a maximum turning point at the point \(A\) and a minimum turning point at the origin \(O\). The line \(l\) touches the curve \(C\) at the point \(A\) and cuts the curve \(C\) at the point \(B\). The \(x\) coordinate of \(A\) is - 4 and the \(x\) coordinate of \(B\) is 2 . The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\) and the line \(l\).
Use integration to find the area of the finite region \(R\).

10. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation $$y = 4 x ^ { 3 } + 9 x ^ { 2 } - 30 x - 8 , \quad - 0.5 \leqslant x \leqslant 2.2$$ The curve has a turning point at the point \(A\).

1. Using calculus, show that the \(x\) coordinate of \(A\) is 1 The curve crosses the \(x\)-axis at the points \(B ( 2,0 )\) and \(C \left( - \frac { 1 } { 4 } , 0 \right)\) The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the line \(A B\), and the \(x\)-axis.
2. Use integration to find the area of the finite region \(R\), giving your answer to 2 decimal places.

8. \begin{figure}[h] \end{figure} The line with equation \(y = x + 5\) cuts the curve with equation \(y = x ^ { 2 } - 3 x + 8\) at the points \(A\) and \(B\), as shown in Fig. 2.

1. Find the coordinates of the points \(A\) and \(B\).
2. Find the area of the shaded region between the curve and the line, as shown in Fig. 2.

14. \begin{figure}[h] \end{figure} Figure 6 shows a sketch of the curve \(C\) with parametric equations $$x = 8 \cos ^ { 3 } \theta , \quad y = 6 \sin ^ { 2 } \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$ Given that the point \(P\) lies on \(C\) and has parameter \(\theta = \frac { \pi } { 3 }\)

1. find the coordinates of \(P\). The line \(l\) is the normal to \(C\) at \(P\).
2. Show that an equation of \(l\) is \(y = x + 3.5\) The finite region \(S\), shown shaded in Figure 6, is bounded by the curve \(C\), the line \(l\), the \(y\)-axis and the \(x\)-axis.
3. Show that the area of \(S\) is given by $$4 + 144 \int _ { 0 } ^ { \frac { \pi } { 3 } } \left( \sin \theta \cos ^ { 2 } \theta - \sin \theta \cos ^ { 4 } \theta \right) d \theta$$
4. Hence, by integration, find the exact area of \(S\).
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

   END

7. \begin{figure}[h] \end{figure} The curves shown in Figure 1 have equations $$y = 6 \cosh x \text { and } y = 9 - 2 \sinh x$$

1. Using the definitions of \(\sinh x\) and \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\), find exact values for the \(x\)-coordinates of the two points where the curves intersect. The finite region between the two curves is shown shaded in Figure 1.
2. Using calculus, find the area of the shaded region, giving your answer in the form \(a \ln b + c\), where \(a , b\) and \(c\) are integers.

4 \includegraphics[max width=\textwidth, alt={}, center]{608720b6-5b18-45e9-8838-c94b347ab3b7-2_547_511_1813_817} The diagram shows a sketch of parts of the curves \(y = \frac { 16 } { x ^ { 2 } }\) and \(y = 17 - x ^ { 2 }\).

1. Verify that these curves intersect at the points \(( 1,16 )\) and \(( 4,1 )\).
2. Calculate the exact area of the shaded region between the curves.

4 \includegraphics[max width=\textwidth, alt={}, center]{367db494-294e-4b53-b9e8-fd2a69fb6069-2_634_670_1123_740} The diagram shows the curve \(y = 4 - x ^ { 2 }\) and the line \(y = x + 2\).

1. Find the \(x\)-coordinates of the points of intersection of the curve and the line.
2. Use integration to find the area of the shaded region bounded by the line and the curve.

10 The equation of a curve is \(y = 7 + 6 x - x ^ { 2 }\).

1. Use calculus to find the coordinates of the turning point on this curve. Find also the coordinates of the points of intersection of this curve with the axes, and sketch the curve.
2. Find \(\int _ { 1 } ^ { 5 } \left( 7 + 6 x - x ^ { 2 } \right) \mathrm { d } x\), showing your working.
3. The curve and the line \(y = 12\) intersect at ( 1,12 ) and ( 5,12 ). Using your answer to part (ii), find the area of the finite region between the curve and the line \(y = 12\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{15b8f97b-c058-409f-907f-cb0a6102abc4-5_643_1034_331_513} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

9 Fig. 9 shows \(P \quad\) The line \(y = x\) \(Q\) The curve \(y = \sqrt { \frac { 1 } { 2 } \left( x + x ^ { 2 } \right) }\) \(R \quad\) The curve \(\quad y = \sqrt { x }\). \begin{figure}[h] \end{figure}

1. Write down the area of the triangle formed by the line \(y = x\), the line \(x = 1\) and the \(x\)-axis.
2. Show that the area of the region formed by the curve \(y = \sqrt { x }\), the line \(x = 1\) and the \(x\)-axis is \(\frac { 2 } { 3 }\). An estimate is required of the Area, \(A\), of the region formed by the curve \(y = \sqrt { \frac { 1 } { 2 } \left( x + x ^ { 2 } \right) }\), the line \(x = 1\) and the \(x\)-axis.
3. Use results to parts (i) and (ii) to complete the statement $$\ldots \ldots \ldots \ldots . . < A < \ldots \ldots \ldots \ldots \ldots . .$$
4. Use the Trapezium Rule with 4 strips to find an estimate for \(A\).
5. Draw a sketch of Fig. 9. Use it to illustrate the area found as the trapezium rule estimate for \(A\).
   Explain how your diagram shows that the trapezium rule estimate must be:
   consistent with the answer to part (iv);
   an under-estimate for A .

5 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-2_486_746_1978_696} The diagram shows the curves \(y = ( 1 - 2 x ) ^ { 5 }\) and \(y = \mathrm { e } ^ { 2 x - 1 } - 1\). The curves meet at the point \(\left( \frac { 1 } { 2 } , 0 \right)\). Find the exact area of the region (shaded in the diagram) bounded by the \(y\)-axis and by part of each curve.

5 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 3 .
2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).

7. \begin{figure}[h] \end{figure} Figure 3 shows part of the curve \(C\) with equation \(y = x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - 34 x\) and the line \(L\) with equation \(y = m x + c\) . The line \(L\) touches \(C\) at the points \(P\) and \(Q\) with \(x\) coordinates \(p\) and \(q\) respectively.

1. Explain why $$x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - ( 34 + m ) x - c = ( x - p ) ^ { 2 } ( x - q ) ^ { 2 }$$ The finite region \(R\) ,shown shaded in Figure 3,is bounded by \(C\) and \(L\) .
2. Use integration by parts to show that the area of \(R\) is \(\frac { ( q - p ) ^ { 5 } } { 30 }\)
3. Show that $$( x - p ) ^ { 2 } ( x - q ) ^ { 2 } = x ^ { 4 } - 2 ( p + q ) x ^ { 3 } + S x ^ { 2 } - T x + U$$ where \(S , T\) and \(U\) are expressions to be found in terms of \(p\) and \(q\) .
4. Using part(a)and part(c)find the value of \(p\) ,the value of \(q\) and the equation of \(L\) .

5 \includegraphics[max width=\textwidth, alt={}, center]{9362eb16-88c9-4279-97aa-907b4916b965-3_646_839_255_653} The diagram shows parts of the curves \(y = x ^ { 2 } + 1\) and \(y = 11 - \frac { 9 } { x ^ { 2 } }\), which intersect at \(( 1,2 )\) and \(( 3,10 )\). Use integration to find the exact area of the shaded region enclosed between the two curves.