1.08d Evaluate definite integrals: between limits

10 At the point \(( 4 , - 1 )\) on a curve, the gradient of the curve is \(- \frac { 3 } { 2 }\). It is given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { - \frac { 1 } { 2 } } + k\), where \(k\) is a constant.

1. Show that \(k = - 2\).
2. Find the equation of the curve.
3. Find the coordinates of the stationary point.
4. Determine the nature of the stationary point.

1 Find the exact value of \(\int _ { 3 } ^ { \infty } \frac { 2 } { x ^ { 2 } } d x\).

3 A curve is such that \(\frac { \mathrm { dy } } { \mathrm { dx } } = 3 ( 4 \mathrm { x } + 5 ) ^ { \frac { 1 } { 2 } }\). It is given that the points \(( 1,9 )\) and \(( 5 , a )\) lie on the curve. Find the value of \(a\).

2 The equation of a curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { ( x - 3 ) ^ { 2 } } + x\). It is given that the curve passes through the point (2, 7). Find the equation of the curve.

11 \includegraphics[max width=\textwidth, alt={}, center]{10b2ec29-adca-4313-ae24-bab8b2d9f8a4-16_505_1166_258_486} The diagram shows the line \(x = \frac { 5 } { 2 }\), part of the curve \(y = \frac { 1 } { 2 } x + \frac { 7 } { 10 } - \frac { 1 } { ( x - 2 ) ^ { \frac { 1 } { 3 } } }\) and the normal to the curve at the point \(A \left( 3 , \frac { 6 } { 5 } \right)\).

1. Find the \(x\)-coordinate of the point where the normal to the curve meets the \(x\)-axis.
2. Find the area of the shaded region, giving your answer correct to 2 decimal places.

1 A curve is such that its gradient at a point \(( x , y )\) is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x - 3 x ^ { - \frac { 1 } { 2 } }\). It is given that the curve passes through the point \(( 4,1 )\). Find the equation of the curve.

11 \includegraphics[max width=\textwidth, alt={}, center]{88c7a3f3-e129-4e9c-acf8-8c96d2668d43-14_693_782_267_669} The diagram shows part of the curve with equation \(y = x + \frac { 2 } { ( 2 x - 1 ) ^ { 2 } }\). The lines \(x = 1\) and \(x = 2\) intersect the curve at \(P\) and \(Q\) respectively and \(R\) is the stationary point on the curve.

1. Verify that the \(x\)-coordinate of \(R\) is \(\frac { 3 } { 2 }\) and find the \(y\)-coordinate of \(R\).
2. Find the exact value of the area of the shaded region.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

9 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 } { ( 2 x + 1 ) ^ { 2 } }\) and \(P ( 1,5 )\) is a point on the curve.

1. The normal to the curve at \(P\) crosses the \(x\)-axis at \(Q\). Find the coordinates of \(Q\).
2. Find the equation of the curve.
3. A point is moving along the curve in such a way that the \(x\)-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the \(y\)-coordinate when \(x = 1\).

9 \includegraphics[max width=\textwidth, alt={}, center]{3b527397-7781-41e9-8218-57277cc977bf-3_391_595_1978_774} The diagram shows part of the curve \(y = \frac { 6 } { 3 x - 2 }\).

1. Find the gradient of the curve at the point where \(x = 2\).
2. Find the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis, giving your answer in terms of \(\pi\).

7 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { ( 1 + 2 x ) ^ { 2 } }\) and the point \(\left( 1 , \frac { 1 } { 2 } \right)\) lies on the curve.

1. Find the equation of the curve.
2. Find the set of values of \(x\) for which the gradient of the curve is less than \(\frac { 1 } { 3 }\).

4

1. Differentiate \(\frac { 2 x ^ { 3 } + 5 } { x }\) with respect to \(x\).
2. Find \(\int ( 3 x - 2 ) ^ { 5 } \mathrm {~d} x\) and hence find the value of \(\int _ { 0 } ^ { 1 } ( 3 x - 2 ) ^ { 5 } \mathrm {~d} x\).

11 \includegraphics[max width=\textwidth, alt={}, center]{4d8fcc3d-a2da-4d98-8500-075d10847be3-4_636_951_255_596} The diagram shows the line \(y = 1\) and part of the curve \(y = \frac { 2 } { \sqrt { } ( x + 1 ) }\).

1. Show that the equation \(y = \frac { 2 } { \sqrt { } ( x + 1 ) }\) can be written in the form \(x = \frac { 4 } { y ^ { 2 } } - 1\).
2. Find \(\int \left( \frac { 4 } { y ^ { 2 } } - 1 \right) \mathrm { d } y\). Hence find the area of the shaded region.
3. The shaded region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { y }\)-axis. Find the exact value of the volume of revolution obtained.

9 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { } ( 4 x + 1 )\) and \(( 2,5 )\) is a point on the curve.

1. Find the equation of the curve.
2. A point \(P\) moves along the curve in such a way that the \(y\)-coordinate is increasing at a constant rate of 0.06 units per second. Find the rate of change of the \(x\)-coordinate when \(P\) passes through \(( 2,5 )\).
3. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \times \frac { \mathrm { d } y } { \mathrm {~d} x }\) is constant.

4 A curve with equation \(y = \mathrm { f } ( x )\) passes through the point \(A ( 3,1 )\) and crosses the \(y\)-axis at \(B\). It is given that \(\mathrm { f } ^ { \prime } ( x ) = ( 3 x - 1 ) ^ { - \frac { 1 } { 3 } }\). Find the \(y\)-coordinate of \(B\).

3 The equation of a curve is \(y = \frac { 2 } { \sqrt { } ( 5 x - 6 ) }\).

1. Find the gradient of the curve at the point where \(x = 2\).
2. Find \(\int \frac { 2 } { \sqrt { } ( 5 x - 6 ) } \mathrm { d } x\) and hence evaluate \(\int _ { 2 } ^ { 3 } \frac { 2 } { \sqrt { } ( 5 x - 6 ) } \mathrm { d } x\).

10 \includegraphics[max width=\textwidth, alt={}, center]{5201a3d5-7733-4d10-9de5-0c2255e3ad60-18_401_584_264_776} The diagram shows part of the curve \(y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)\), defined for \(x \geqslant 0\).

1. Find, showing all necessary working, the area of the shaded region.
2. Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
3. Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis.

11 \includegraphics[max width=\textwidth, alt={}, center]{097c5d00-9f92-4c3e-8056-7de09347fbb6-18_515_853_260_644} The diagram shows part of the curve \(y = ( 1 + 4 x ) ^ { \frac { 1 } { 2 } }\) and a point \(P ( 6,5 )\) lying on the curve. The line \(P Q\) intersects the \(x\)-axis at \(Q ( 8,0 )\).

1. Show that \(P Q\) is a normal to the curve.
2. Find, showing all necessary working, the exact volume of revolution obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
   [0pt] [In part (ii) you may find it useful to apply the fact that the volume, \(V\), of a cone of base radius \(r\) and vertical height \(h\), is given by \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]

7

1. Find the quotient when \(9 x ^ { 3 } - 6 x ^ { 2 } - 20 x + 1\) is divided by ( \(3 x + 2\) ), and show that the remainder is 9 .
2. Hence find \(\int _ { 1 } ^ { 6 } \frac { 9 x ^ { 3 } - 6 x ^ { 2 } - 20 x + 1 } { 3 x + 2 } \mathrm {~d} x\), giving the answer in the form \(a + \ln b\) where \(a\) and \(b\) are integers.
3. Find the exact root of the equation \(9 e ^ { 9 y } - 6 e ^ { 6 y } - 20 e ^ { 3 y } - 8 = 0\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7 It is given that \(\int _ { 0 } ^ { a } \left( \frac { 4 } { 2 x + 1 } + 8 x \right) \mathrm { d } x = 10\), where \(a\) is a positive constant.

1. Show that \(a = \sqrt { 2.5 - 0.5 \ln ( 2 a + 1 ) }\).
2. Using the equation in part (a), show by calculation that \(1 < a < 2\).
3. Use an iterative formula, based on the equation in part (a), to find the value of \(a\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4

1. Use the trapezium rule with three intervals to show that the value of \(\int _ { 1 } ^ { 4 } \ln x \mathrm {~d} x\) is approximately \(\ln 12\).
2. Use a graph of \(y = \ln x\) to show that \(\ln 12\) is an under-estimate of the true value of \(\int _ { 1 } ^ { 4 } \ln x \mathrm {~d} x\).

6 \includegraphics[max width=\textwidth, alt={}, center]{1b9c6b41-69dd-4132-92c7-9507cbd741dd-08_542_661_269_731} The diagram shows the curve \(y = 3 \mathrm { e } ^ { 2 x - 1 }\). The shaded region is bounded by the curve and the lines \(x = a , x = a + 1\) and \(y = 0\), where \(a\) is a constant. It is given that the area of the shaded region is 120 square units.

1. Show that \(a = \frac { 1 } { 2 } \ln \left( 80 + \mathrm { e } ^ { 2 a - 1 } \right) - \frac { 1 } { 2 }\).
2. Use an iterative formula, based on the equation in part (a), to find the value of \(a\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

6 \includegraphics[max width=\textwidth, alt={}, center]{712be8e6-e1e9-4662-b1f1-51c39c2c9df1-08_542_661_269_731} The diagram shows the curve \(y = 3 \mathrm { e } ^ { 2 x - 1 }\). The shaded region is bounded by the curve and the lines \(x = a , x = a + 1\) and \(y = 0\), where \(a\) is a constant. It is given that the area of the shaded region is 120 square units.

1. Show that \(a = \frac { 1 } { 2 } \ln \left( 80 + \mathrm { e } ^ { 2 a - 1 } \right) - \frac { 1 } { 2 }\).
2. Use an iterative formula, based on the equation in part (a), to find the value of \(a\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

3 It is given that \(\int _ { 0 } ^ { a } \left( 3 \mathrm { e } ^ { 2 x } - 1 \right) \mathrm { d } x = 12\), where \(a\) is a positive constant.

1. Show that \(a = \frac { 1 } { 2 } \ln \left( 9 + \frac { 2 } { 3 } a \right)\).
2. Use an iterative formula, based on the equation in (a), to find the value of \(a\) correct to 4 significant figures. Use an initial value of 1 and give the result of each iteration to 6 significant figures. [3]

3 \includegraphics[max width=\textwidth, alt={}, center]{4ce3208e-8ceb-4848-a9c7-fcda166319f4-04_458_892_269_614} The diagram shows part of the curve \(y = \frac { 6 } { 2 x + 3 }\). The shaded region is bounded by the curve and the lines \(x = 6\) and \(y = 2\). Find the exact area of the shaded region, giving your answer in the form \(a - \ln b\), where \(a\) and \(b\) are integers.

3 \includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-04_458_892_269_614} The diagram shows part of the curve \(y = \frac { 6 } { 2 x + 3 }\). The shaded region is bounded by the curve and the lines \(x = 6\) and \(y = 2\). Find the exact area of the shaded region, giving your answer in the form \(a - \ln b\), where \(a\) and \(b\) are integers.