1.08e Area between curve and x-axis: using definite integrals

4 \includegraphics[max width=\textwidth, alt={}, center]{4d03f4e3-ae6c-4a0d-ae4d-d89258d2919a-3_588_1136_255_502} The diagram shows the curve \(y = - 1 + \sqrt { x + 4 }\) and the line \(y = 3\).

1. Show that \(y = - 1 + \sqrt { x + 4 }\) can be rearranged as \(x = y ^ { 2 } + 2 y - 3\).
2. Hence find by integration the exact area of the shaded region enclosed between the curve, the \(y\)-axis and the line \(y = 3\).

7 \includegraphics[max width=\textwidth, alt={}, center]{b2c1188d-881e-4fb5-bece-5a51006543c7-3_519_611_1087_712} The diagram shows the curve \(y = x ^ { \frac { 3 } { 2 } } - 1\), which crosses the \(x\)-axis at \(( 1,0 )\), and the tangent to the curve at the point \(( 4,7 )\).

1. Show that \(\int _ { 1 } ^ { 4 } \left( x ^ { \frac { 3 } { 2 } } - 1 \right) \mathrm { d } x = 9 \frac { 2 } { 5 }\).
2. Hence find the exact area of the shaded region enclosed by the curve, the tangent and the \(x\)-axis.

7 The cubic polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = x ^ { 3 } - 3 x ^ { 2 } - x + 3\).

1. Find the quotient and remainder when \(\mathrm { f } ( x )\) is divided by \(( x + 1 )\).
2. Hence find the three roots of the equation \(\mathrm { f } ( x ) = 0\). \includegraphics[max width=\textwidth, alt={}, center]{555f7205-5e2a-4471-901d-d8abc9dd4b4a-3_540_718_1466_660} The diagram shows the curve \(C\) with equation \(y = x ^ { 4 } - 4 x ^ { 3 } - 2 x ^ { 2 } + 12 x + 9\).
3. Show that the \(x\)-coordinates of the stationary points on \(C\) are given by \(x ^ { 3 } - 3 x ^ { 2 } - x + 3 = 0\).
4. Use integration to find the exact area of the region enclosed by \(C\) and the \(x\)-axis.

10 Fig. 10 shows a sketch of the graph of \(y = 7 x - x ^ { 2 } - 6\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the equation of the tangent to the curve at the point on the curve where \(x = 2\). Show that this tangent crosses the \(x\)-axis where \(x = \frac { 2 } { 3 }\).
2. Show that the curve crosses the \(x\)-axis where \(x = 1\) and find the \(x\)-coordinate of the other point of intersection of the curve with the \(x\)-axis.
3. Find \(\int _ { 1 } ^ { 2 } \left( 7 x - x ^ { 2 } - 6 \right) \mathrm { d } x\). Hence find the area of the region bounded by the curve, the tangent and the \(x\)-axis, shown shaded on Fig. 10.

12 The equation of a curve is \(y = 9 x ^ { 2 } - x ^ { 4 }\).

1. Show that the curve meets the \(x\)-axis at the origin and at \(x = \pm a\), stating the value of \(a\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\). Hence show that the origin is a minimum point on the curve. Find the \(x\)-coordinates of the maximum points.
3. Use calculus to find the area of the region bounded by the curve and the \(x\)-axis between \(x = 0\) and \(x = a\), using the value you found for \(a\) in part (i).

13 Fig. 13.1 shows a greenhouse which is built against a wall. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} The greenhouse is a prism of length 5.5 m . The curve AC is an arc of a circle with centre B and radius 2.1 m , as shown in Fig. 13.2. The sector angle ABC is 1.8 radians and ABD is a straight line. The curved surface of the greenhouse is covered in polythene.

1. Find the length of the arc AC and hence find the area of polythene required for the curved surface of the greenhouse.
2. Calculate the length BD .
3. Calculate the volume of the greenhouse.

5 \includegraphics[max width=\textwidth, alt={}, center]{774bb427-5392-45d3-8e4e-47d08fb8a792-02_559_1191_1749_479} The diagram shows the curve with equation \(y = \frac { 6 } { \sqrt { 3 x - 2 } }\). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 1 , x = a\) and \(y = 0\), where \(a\) is a constant greater than 1 . It is given that the area of \(R\) is 16 square units. Find the value of \(a\) and hence find the exact volume of the solid formed when \(R\) is rotated completely about the \(x\)-axis.
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6 \includegraphics[max width=\textwidth, alt={}, center]{fc7679bf-a9a1-493d-bf89-35206382787f-3_576_821_258_662} The diagram shows the curve with equation \(y = \sqrt { 3 x - 5 }\). The tangent to the curve at the point \(P\) passes through the origin. The shaded region is bounded by the curve, the \(x\)-axis and the line \(O P\). Show that the \(x\)-coordinate of \(P\) is \(\frac { 10 } { 3 }\) and hence find the exact area of the shaded region.

9 \includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-4_661_915_269_557} The diagram shows the curve $$y = \mathrm { e } ^ { 2 x } - 18 x + 15 .$$ The curve crosses the \(y\)-axis at \(P\) and the minimum point is \(Q\). The shaded region is bounded by the curve and the line \(P Q\).

1. Show that the \(x\)-coordinate of \(Q\) is \(\ln 3\).
2. Find the exact area of the shaded region.

7 \includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-3_547_851_1749_605} The diagram shows the curve \(y = \sqrt { \frac { 3 } { 4 x + 1 } }\) for \(0 \leqslant x \leqslant 20\). The point \(P\) on the curve has coordinates \(\left( 20 , \frac { 1 } { 9 } \sqrt { 3 } \right)\). The shaded region \(R\) is enclosed by the curve and the lines \(x = 0\) and \(y = \frac { 1 } { 9 } \sqrt { 3 }\).

1. Find the exact area of \(R\).
2. Find the exact volume of the solid obtained when \(R\) is rotated completely about the \(x\)-axis.

5 \includegraphics[max width=\textwidth, alt={}, center]{00a4be37-c095-4d9c-a1cd-d03b8ab1d411-2_455_643_1327_694} The diagram shows the curve \(y = \mathrm { e } ^ { 3 x } - 6 \mathrm { e } ^ { 2 x } + 32\).

1. Find the exact \(x\)-coordinate of the minimum point and verify that the \(y\)-coordinate of the minimum point is 0 .
2. Find the exact area of the region (shaded in the diagram) enclosed by the curve and the axes.

5 \includegraphics[max width=\textwidth, alt={}, center]{6d15cb4d-f540-488b-b94e-7a494f192ba5-2_469_721_1932_662} The diagram shows the curves \(y = \mathrm { e } ^ { 2 x }\) and \(y = 8 \mathrm { e } ^ { - x }\). The shaded region is bounded by the curves and the \(y\)-axis. Without using a calculator,

1. solve an appropriate equation to show that the curves intersect at a point for which \(x = \ln 2\),
2. find the area of the shaded region, giving your answer in simplified form.

8 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h] \end{figure}

1. Find the gradient of PR.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
3. Find the exact coordinates of the turning point Q .
4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

8 Fig. 8 shows the line \(y = x\) and parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where $$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$ The curves intersect the axes at the points A and B , as shown. The curves and the line \(y = x\) meet at the point C . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of A and B . Verify that the coordinates of C are \(( 1,1 )\).
2. Prove algebraically that \(\mathrm { g } ( x )\) is the inverse of \(\mathrm { f } ( x )\).
3. Evaluate \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\), giving your answer in terms of e.
4. Use integration by parts to find \(\int \ln x \mathrm {~d} x\). Hence show that \(\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }\).
5. Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .

8 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }\), with its turning point P . \begin{figure}[h] \end{figure}

1. Write down the coordinates of the intercepts of \(y = \mathrm { f } ( x )\) with the \(x\) - and \(y\)-axes.
2. Find the exact coordinates of the turning point P .
3. Show that the exact area of the region enclosed by the curve and the \(x\) - and \(y\)-axes is \(\frac { 1 } { 4 } \left( e ^ { 2 } - 3 \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\).
4. Express \(\mathrm { g } ( x )\) in terms of \(x\). Sketch the curve \(y = \mathrm { g } ( x )\) on the copy of Fig. 8, indicating the coordinates of its intercepts with the \(x\) - and \(y\)-axes and of its turning point.
5. Write down the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\) and the \(x\)-and \(y\)-axes.

8 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } - 1 , x \in \mathbb { R } .$$ The curve crosses the \(x\)-axis at O and P , and has a turning point at Q . \begin{figure}[h] \end{figure}

1. Find the exact \(x\)-coordinate of P .
2. Show that the \(x\)-coordinate of Q is \(\ln 2\) and find its \(y\)-coordinate.
3. Find the exact area of the region enclosed by the curve and the \(x\)-axis. The domain of \(\mathrm { f } ( x )\) is now restricted to \(x \geqslant \ln 2\).
4. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Write down its domain and range, and sketch its graph on the copy of Fig. 9.

9 \includegraphics[max width=\textwidth, alt={}, center]{074597e7-5bb1-4249-9cfa-784974a6fd2b-4_486_1097_696_523} The diagram shows the curve with equation \(y = \sqrt { 2 x + 1 }\) between the points \(A \left( - \frac { 1 } { 2 } , 0 \right)\) and \(B ( 4,3 )\).

1. Find the area of the region bounded by the curve, the \(x\)-axis and the line \(x = 4\). Hence find the area of the region bounded by the curve and the lines \(O A\) and \(O B\), where \(O\) is the origin.
2. Show that the curve between \(B\) and \(A\) can be expressed in polar coordinates as $$r = \frac { 1 } { 1 - \cos \theta } , \quad \text { where } \tan ^ { - 1 } \left( \frac { 3 } { 4 } \right) \leqslant \theta \leqslant \pi$$
3. Deduce from parts (i) and (ii) that \(\int _ { \tan ^ { - 1 } \left( \frac { 3 } { 4 } \right) } ^ { \pi } \operatorname { cosec } ^ { 4 } \left( \frac { 1 } { 2 } \theta \right) \mathrm { d } \theta = 24\). www.ocr.org.uk after the live examination series.
   If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1 GE.
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11

1. Use the substitution \(u ^ { 2 } = x ^ { 2 } + 3\) to show that \(\int \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 3 } } \mathrm {~d} x = \frac { 4 } { 3 } \left( x ^ { 2 } - 6 \right) \sqrt { x ^ { 2 } + 3 } + c\).
2. In this question you must show detailed reasoning. \includegraphics[max width=\textwidth, alt={}, center]{6b766f5c-8533-4e0c-bb10-0d9949dc777b-7_620_951_1836_317} The graph shows part of the curve \(y = \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 2 } }\).
   Find the exact area enclosed by the curve \(y = \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 3 } }\), the normal to this curve at the point \(( 1,2 )\) and the \(x\)-axis.

5 The diagram shows the circle with centre O and radius 2, and the parabola \(y = \frac { 1 } { \sqrt { 3 } } \left( 4 - x ^ { 2 } \right)\). \includegraphics[max width=\textwidth, alt={}, center]{f2f45d6c-cfdc-455b-ab08-597b06a69f36-06_838_970_1059_280} The circle meets the parabola at points \(P\) and \(Q\), as shown in the diagram.

1. Verify that the coordinates of \(Q\) are \(( 1 , \sqrt { 3 } )\).
2. Find the exact area of the shaded region enclosed by the \(\operatorname { arc } P Q\) of the circle and the parabola.
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15. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 32 } { x ^ { 2 } } + 3 x - 8 , \quad x > 0$$ The point \(P ( 4,6 )\) lies on \(C\).
The line \(l\) is the normal to \(C\) at the point \(P\).
The region \(R\), shown shaded in Figure 4, is bounded by the line \(l\), the curve \(C\), the line with equation \(x = 2\) and the \(x\)-axis. Show that the area of \(R\) is 46
(Solutions based entirely on graphical or numerical methods are not acceptable.)

13. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation $$y = 2 x ^ { 3 } - 17 x ^ { 2 } + 40 x$$ The curve has a minimum turning point at \(x = k\).
The region \(R\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the line with equation \(x = k\). Show that the area of \(R\) is \(\frac { 256 } { 3 }\) (Solutions based entirely on graphical or numerical methods are not acceptable.)

10. $$g ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 41 x - 70$$

1. Use the factor theorem to show that \(\mathrm { g } ( x )\) is divisible by \(( x - 5 )\).
2. Hence, showing all your working, write \(\mathrm { g } ( x )\) as a product of three linear factors. The finite region \(R\) is bounded by the curve with equation \(y = \mathrm { g } ( x )\) and the \(x\)-axis, and lies below the \(x\)-axis.
3. Find, using algebraic integration, the exact value of the area of \(R\).

10. \begin{figure}[h] \end{figure} In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
Figure 2 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 1 } { 3 } x ^ { 2 } - 2 \sqrt { x } + 3 \quad x \geqslant 0$$ The point \(P\) lies on \(C\) and has \(x\) coordinate 4
The line \(l\) is the tangent to \(C\) at \(P\).

1. Show that \(l\) has equation $$13 x - 6 y - 26 = 0$$ The region \(R\), shown shaded in Figure 2, is bounded by the \(y\)-axis, the curve \(C\), the line \(l\) and the \(x\)-axis.
2. Find the exact area of \(R\).

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

Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} The finite region \(R\), shown shaded in Figure 2, is bounded by the curve with equation \(y = 4 x ^ { 2 } + 3\), the \(y\)-axis and the line with equation \(y = 23\) Show that the exact area of \(R\) is \(k \sqrt { 5 }\) where \(k\) is a rational constant to be found.