1.08a Fundamental theorem of calculus: integration as reverse of differentiation

1. A curve has equation \(y = \mathrm { f } ( x ) , x \geqslant 0\)

Given that

• \(\mathrm { f } ^ { \prime } ( x ) = 4 x + a \sqrt { x } + b\), where \(a\) and \(b\) are constants
• the curve has a stationary point at \(( 4,3 )\)
• the curve meets the \(y\)-axis at - 5
  find \(\mathrm { f } ( x )\), giving your answer in simplest form.

15. \begin{figure}[h] \end{figure} Figure 7 shows an open tank for storing water, \(A B C D E F\). The sides \(A C D F\) and \(A B E F\) are rectangles. The faces \(A B C\) and \(F E D\) are sectors of a circle with radius \(A B\) and \(F E\) respectively.

• \(A B = F E = r \mathrm {~cm}\)
• \(A F = B E = C D = l \mathrm {~cm}\)
• angle \(B A C =\) angle \(E F D = 0.9\) radians

Given that the volume of the tank is \(360 \mathrm {~cm} ^ { 3 }\) a. show that the surface area of the tank, \(S \mathrm {~cm} ^ { 2 }\), is given by $$S = 0.9 r ^ { 2 } + \frac { 1600 } { r }$$ (4) Given that \(r\) can vary
b. use calculus to find the value of \(r\) for which \(S\) is stationary.
c. Find, to 3 significant figures the minimum value of \(S\).

7. A curve \(C\) has equation \(y = \mathrm { f } ( x )\). Given that

• \(\mathrm { f } ^ { \prime } ( x ) = 18 x ^ { 2 } + 2 a x + b\)
• the \(y\)-intercept of \(C\) is - 48
• the point \(A\), with coordinates \(( - 1,45 )\) lies on \(C\) a. Show that \(a - b = 99\) b. Find the value of \(a\) and the value of \(b\).
  c. Show that \(( 2 x + 1 )\) is a factor of \(\mathrm { f } ( x )\).

1. The curve \(C\) has equation \(y = \mathrm { f } ( x )\)

The curve

• passes through the point \(P ( 3 , - 10 )\)
• has a turning point at \(P\)

Given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x ^ { 3 } - 9 x ^ { 2 } + 5 x + k$$ where \(k\) is a constant,

1. show that \(k = 12\)
2. Hence find the coordinates of the point where \(C\) crosses the \(y\)-axis.

1. A curve \(C\) has equation \(y = \mathrm { f } ( x )\)

Given that

• \(\mathrm { f } ^ { \prime } ( x ) = 6 x ^ { 2 } + a x - 23\) where \(a\) is a constant
• the \(y\) intercept of \(C\) is - 12
• ( \(x + 4\) ) is a factor of \(\mathrm { f } ( x )\) find, in simplest form, \(\mathrm { f } ( x )\)

5 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } - 3 x\). The curve passes through the point (6, 20).

1. Determine the equation of the curve.
2. Hence determine \(\int _ { 1 } ^ { p } y \mathrm {~d} x\) in terms of the constant \(p\).

6 A curve \(C\) has an equation which satisfies \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 3 x ^ { 2 } + 2\), for all values of \(x\).

1. It is given that \(C\) has a single stationary point. Determine the nature of this stationary point. The diagram shows the graph of the gradient function for \(C\). \includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-04_702_442_1672_242}
2. Given that \(C\) passes through the point \(\left( - 1 , \frac { 1 } { 4 } \right)\), find the equation of \(C\) in the form \(y = \mathrm { f } ( x )\).

8

1. The quadratic polynomial \(a x ^ { 2 } + b x\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { f } ( x )\).
   Use differentiation from first principles to determine, in terms of \(a , b\) and \(x\), an expression for \(\mathrm { f } ^ { \prime } ( x )\).
2. \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-07_565_1043_516_317} $$y = a x ^ { 2 } + b x$$ The diagram shows the quadratic curve \(y = a x ^ { 2 } + b x\), where \(a\) and \(b\) are constants. The shaded region is enclosed by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 4\). The tangent to the curve at \(x = 4\) intersects the \(x\)-axis at the point with coordinates \(( k , 0 )\).
   Given that the area of the shaded region is 9 units \({ } ^ { 2 }\), and the gradient of this tangent is \(- \frac { 3 } { 4 }\), determine the value of \(k\).

11 \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-08_586_672_1231_242} A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds, where \(t \geqslant 0\), the velocity of \(P\) in the positive \(x\)-direction is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that \(v = t ( t - 3 ) ( 8 - t )\). \(P\) attains its maximum velocity at time \(T\) seconds. The diagram shows part of the velocity-time graph for the motion of \(P\).

1. State the acceleration of \(P\) at time \(T\).
2. In this question you must show detailed reasoning. Determine the value of \(T\).
3. Find the total distance that \(P\) travels between times \(t = 0\) and \(t = T\). \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-09_524_410_251_242} Particles \(P\) and \(Q\), of masses 4 kg and 6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley. The system is in equilibrium with \(P\) hanging 1.75 m above a horizontal plane and \(Q\) resting on the plane. Both parts of the string below the pulley are vertical (see diagram).
   1. Find the magnitude of the normal reaction force acting on \(Q\). The mass of \(P\) is doubled, and the system is released from rest. You may assume that in the subsequent motion \(Q\) does not reach the pulley.
   2. Determine the magnitude of the force exerted on the pulley by the string before \(P\) strikes the plane.
   3. Determine the total distance travelled by \(Q\) between the instant when the system is released and the instant when \(Q\) first comes momentarily to rest. When this motion is observed in practice, it is found that the total distance travelled by \(Q\) between the instant when the system is released and the instant when \(Q\) first comes momentarily to rest is less than the answer calculated in part (c).
   4. State one factor that could account for this difference.

10 A particle \(P\) is moving in a straight line. At time \(t\) seconds, where \(t \geqslant 0 , P\) has velocity \(v \mathrm {~ms} ^ { - 1 }\) and acceleration \(a \mathrm {~ms} ^ { - 2 }\) where \(a = 4 t - 9\). It is given that \(v = 2\) when \(t = 1\).

1. Find an expression for \(v\) in terms of \(t\). The particle \(P\) is instantaneously at rest when \(t = t _ { 1 }\) and \(t = t _ { 2 }\), where \(t _ { 1 } < t _ { 2 }\).
2. Find the values of \(t _ { 1 }\) and \(t _ { 2 }\).
3. Determine the total distance travelled by \(P\) between times \(t = 0\) and \(t = t _ { 2 }\).

6 The gradient of a curve is given by the equation \(\frac { d y } { d x } = 6 x ^ { 2 } - 20 x + 6\). The curve passes through the point \(( 2,6 )\).

1. Find the equation of the curve.
2. Verify that the equation of the curve can be written as \(y = 2 ( x + 1 ) ( x - 3 ) ^ { 2 }\).
3. Sketch the curve, indicating the points where the curve meets the axes.

12 The diagram shows the graph of \(\mathrm { f } ( \mathrm { x } ) = \mathrm { k } ( \mathrm { x } - \mathrm { p } ) ( \mathrm { x } - \mathrm { q } )\) where \(k , p\) and \(q\) are constants. The graph passes through the points \(( - 1,0 ) , ( 0 , - 4 )\) and \(( 2,0 )\). \includegraphics[max width=\textwidth, alt={}, center]{b5c47a93-ce43-4aa1-ba7f-fbb650523373-7_775_638_347_242}

1. Find \(\mathrm { f } ( \mathrm { x } )\) in the form \(\mathrm { ax } ^ { 2 } + \mathrm { bx } + \mathrm { c }\). A cubic curve has gradient function \(f ( x )\). This cubic curve passes through the point \(( 0,8 )\).
2. Find the equation of the cubic curve.
3. Determine the coordinates of the stationary points of the cubic curve.

7 The diagram shows part of a curve which passes through the point \(( 1,0 )\). \includegraphics[max width=\textwidth, alt={}, center]{5428eabf-431d-4db1-8c25-1f2b9570d9aa-4_711_704_1722_258} The gradient of the curve is given by \(\frac { d y } { d x } = 6 x + \frac { 8 } { x ^ { 3 } }\).
Determine whether the curve passes through the point \(( 2,12 )\).

12 A model boat has velocity \(\mathbf { v } = ( ( 2 t - 2 ) \mathbf { i } + ( 2 t + 2 ) \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) for \(t \geq 0\), where \(t\) is the time in seconds. \(\mathbf { i }\) is the unit vector east and \(\mathbf { j }\) is the unit vector north.
When \(t = 3\), the position vector of the boat is \(( 3 \mathbf { i } + 14 \mathbf { j } ) \mathrm { m }\).

1. Show that the boat is never instantaneously at rest.
2. Determine any times at which the boat is moving directly northwards.
3. Determine any times at which the boat is north-east of the origin.

2 Given that \(y = \left( x ^ { 2 } + 5 \right) ^ { 12 }\),

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence find \(\int 48 x \left( x ^ { 2 } + 5 \right) ^ { 11 } \mathrm {~d} x\).

11 In this question you must show detailed reasoning.
The variables \(x\) and \(y\) are such that \(\frac { \mathrm { dy } } { \mathrm { dx } }\) is directly proportional to the square root of \(x\).
When \(x = 4 , \frac { d y } { d x } = 3\).

1. Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) in terms of \(x\). When \(\mathrm { x } = 4 , \mathrm { y } = 10\).
2. Find \(y\) in terms of \(x\).

6 The gradient, \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), of a curve at the point \(( x , y )\) is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 10 x ^ { 4 } - 6 x ^ { 2 } + 5$$ The curve passes through the point \(P ( 1,4 )\).

1. Find the equation of the tangent to the curve at the point \(P\), giving your answer in the form \(y = m x + c\).
2. Find the equation of the curve.

1. The curve with equation \(y = \mathrm { f } ( x )\) passes through the point (8, 7).

Given that $$\mathrm { f } ^ { \prime } ( x ) = 4 x ^ { \frac { 1 } { 3 } } - 5$$ find \(\mathrm { f } ( x )\).

9. The curve \(C\) has the equation \(y = \mathrm { f } ( x )\) where $$f ^ { \prime } ( x ) = 1 + \frac { 2 } { \sqrt { x } } , \quad x > 0$$ The straight line \(l\) has the equation \(y = 2 x - 1\) and is a tangent to \(C\) at the point \(P\).

1. State the gradient of \(C\) at \(P\).
2. Find the \(x\)-coordinate of \(P\).
3. Find an equation for \(C\).
4. Show that \(C\) crosses the \(x\)-axis at the point \(( 1,0 )\) and at no other point.

1. The curve \(y = \mathrm { f } ( x )\) passes through the point \(P ( - 1,3 )\) and is such that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 1 } { x ^ { 2 } } , \quad x \neq 0 .$$

1. Using integration, find \(\mathrm { f } ( x )\).
2. Sketch the curve \(y = \mathrm { f } ( x )\) and write down the equations of its asymptotes.

10. The curve \(C\) has the equation \(y = \mathrm { f } ( x )\). Given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - \frac { 2 } { x ^ { 2 } } , \quad x \neq 0 ,$$ and that the point \(A\) on \(C\) has coordinates (2, 6),

1. find an equation for \(C\),
2. find an equation for the tangent to \(C\) at \(A\), giving your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers,
3. show that the line \(y = x + 3\) is also a tangent to \(C\).

8 A curve, drawn from the origin \(O\), crosses the \(x\)-axis at the point \(A ( 9,0 )\). Tangents to the curve at \(O\) and \(A\) meet at the point \(P\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{48c5470e-6489-4b25-98a6-1b4e101ab01c-006_763_879_466_577} The curve, defined for \(x \geqslant 0\), has equation $$y = x ^ { \frac { 3 } { 2 } } - 3 x$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. 1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(O\) and hence write down an equation of the tangent at \(O\).
   2. Show that the equation of the tangent at \(A ( 9,0 )\) is \(2 y = 3 x - 27\).
   3. Hence find the coordinates of the point \(P\) where the two tangents meet.
3. Find \(\int \left( x ^ { \frac { 3 } { 2 } } - 3 x \right) \mathrm { d } x\).
4. Calculate the area of the shaded region bounded by the curve and the tangents \(O P\) and \(A P\).

4

1. Write \(\sqrt { x }\) in the form \(x ^ { k }\), where \(k\) is a fraction.
2. Hence express \(\sqrt { x } ( x - 1 )\) in the form \(x ^ { p } - x ^ { q }\).
3. Find \(\int \sqrt { x } ( x - 1 ) \mathrm { d } x\).
4. Hence show that \(\int _ { 1 } ^ { 2 } \sqrt { x } ( x - 1 ) \mathrm { d } x = \frac { 4 } { 15 } ( \sqrt { 2 } + 1 )\).

2 At the point \(( x , y )\) on a curve, where \(x > 0\), the gradient is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 7 \sqrt { x ^ { 5 } } - 4$$

1. Write \(\sqrt { x ^ { 5 } }\) in the form \(x ^ { k }\), where \(k\) is a fraction.
2. Find \(\int \left( 7 \sqrt { x ^ { 5 } } - 4 \right) \mathrm { d } x\).
3. Hence find the equation of the curve, given that the curve passes through the point \(( 1,3 )\).

4 A curve is defined for \(x > 0\). The gradient of the curve at the point \(( x , y )\) is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x ^ { 2 } } - \frac { x } { 4 }$$

1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. The curve has a stationary point \(M\) whose \(y\)-coordinate is \(\frac { 5 } { 2 }\).
   1. Find the \(x\)-coordinate of \(M\).
   2. Use your answers to parts (a) and (b)(i) to show that \(M\) is a maximum point.
   3. Find the equation of the curve.