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

9. A curve with equation \(y = \mathrm { f } ( x )\) passes through the point ( 3,6 ). Given that $$f ^ { \prime } ( x ) = ( x - 2 ) ( 3 x + 4 )$$

1. use integration to find \(\mathrm { f } ( x )\). Give your answer as a polynomial in its simplest form.
2. Show that \(\mathrm { f } ( x ) \equiv ( x - 2 ) ^ { 2 } ( x + p )\), where \(p\) is a positive constant. State the value of \(p\).
3. Sketch the graph of \(y = \mathrm { f } ( x )\), showing the coordinates of any points where the curve touches or crosses the coordinate axes.

7. (a) Show that \(\frac { ( 3 - \sqrt { } x ) ^ { 2 } } { \sqrt { } x }\) can be written as \(9 x ^ { - \frac { 1 } { 2 } } - 6 + x ^ { \frac { 1 } { 2 } }\). Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 3 - \sqrt { } x ) ^ { 2 } } { \sqrt { } x } , x > 0\), and that \(y = \frac { 2 } { 3 }\) at \(x = 1\),
(b) find \(y\) in terms of \(x\).

7. \begin{figure}[h] \end{figure} Figure 1 shows a hemispherical bowl.
Water is flowing into the bowl at a constant rate of \(180 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
When the height of the water is \(h \mathrm {~cm}\), the volume of water \(V \mathrm {~cm} ^ { 3 }\) is given by $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 90 - h ) , \quad 0 \leqslant h \leqslant 30$$ Find the rate of change of the height of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 15\) Give your answer to 2 significant figures.

3. \begin{figure}[h] \end{figure} A bowl with circular cross section and height 20 cm is shown in Figure 2.
The bowl is initially empty and water starts flowing into the bowl.
When the depth of water is \(h \mathrm {~cm}\), the volume of water in the bowl, \(V \mathrm {~cm} ^ { 3 }\), is modelled by the equation $$V = \frac { 1 } { 3 } h ^ { 2 } ( h + 4 ) \quad 0 \leqslant h \leqslant 20$$ Given that the water flows into the bowl at a constant rate of \(160 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\), find, according to the model,

1. the time taken to fill the bowl,
2. the rate of change of the depth of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 5\)

A curve has equation \(y = \mathrm { f } ( x )\) and passes through the point (4, 22). Given that $$\mathrm { f } ^ { \prime } ( x ) = 3 x ^ { 2 } - 3 x ^ { \frac { 1 } { 2 } } - 7 ,$$ use integration to find \(\mathrm { f } ( x )\), giving each term in its simplest form.

A curve with equation \(y = \mathrm { f } ( x )\) passes through the point \(( 4,9 )\). Given that $$f ^ { \prime } ( x ) = \frac { 3 \sqrt { } x } { 2 } - \frac { 9 } { 4 \sqrt { } x } + 2 , \quad x > 0$$

1. find \(\mathrm { f } ( x )\), giving each term in its simplest form. Point \(P\) lies on the curve. The normal to the curve at \(P\) is parallel to the line \(2 y + x = 0\)
2. Find the \(x\) coordinate of \(P\).

5 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 12 \sqrt { x }\). The curve passes through the point (4,50). Find the equation of the curve.

9 A curve has gradient given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { 2 } + 8 x\). The curve passes through the point \(( 1,5 )\). Find the equation of the curve.

7 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { x ^ { 3 } }\). The curve passes through \(( 1,4 )\).
Find the equation of the curve.

8 The gradient of a curve is \(3 \sqrt { x } - 5\). The curve passes through the point ( 4,6 ). Find the equation of the curve.

1 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 - 5 x\).
Find the equation of the curve given that it passes through the point \(( 0,1 )\).

5 The gradient of a curve is given by the function \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 - x\).
The curve passes through the point \(( 1,2 )\).
Find the equation of the curve.

7 The gradient of a curve \(y = \mathrm { f } ( x )\) is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 10 x + 6\). The curve passes through the point \(( 2,3 )\) Find the equation of the curve.

9 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 12 x + 9\). The curve passes through the point \(( 2 , - 2 )\).

1. Find the equation of the curve.
2. Show that the curve touches the \(x\)-axis at one point (A) and cuts it at another (B). State the coordinates of A and B.
3. The curve cuts the \(y\)-axis at C . Show that the tangent at C is perpendicular to the normal at B.

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. Given that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 3 } - 4 } { x ^ { 3 } } , \quad x \neq 0$$ and that \(y = 0\) when \(x = - 1\), find the value of \(y\) when \(x = 2\).

9. \includegraphics[max width=\textwidth, alt={}, center]{e4afa57d-5be3-42a6-ab35-39b0fdcc1681-3_559_732_824_388} The diagram shows the curve with equation \(y = \mathrm { f } ( x )\) which crosses the \(x\)-axis at the origin and at the points \(A\) and \(B\). Given that $$f ^ { \prime } ( x ) = 4 - 6 x - 3 x ^ { 2 }$$

1. find an expression for \(y\) in terms of \(x\),
2. show that \(A\) has coordinates ( \(- 4,0\) ) and find the coordinates of \(B\),
3. find the total area of the two regions bounded by the curve and the \(x\)-axis.

6. Given that $$\mathrm { f } ^ { \prime } ( x ) = 5 + \frac { 4 } { x ^ { 2 } } , \quad x \neq 0$$

1. find an expression for \(\mathrm { f } ( x )\). Given also that $$\mathrm { f } ( 2 ) = 2 \mathrm { f } ( 1 ) ,$$
2. find \(\mathrm { f } ( 4 )\).

3 Two girls, Marie and Nina, are members of an Olympic hockey team. They are doing fitness training. Marie runs along a straight line at a constant speed of \(6 \mathrm {~ms} ^ { - 1 }\). Nina is stationary at a point O on the line until Marie passes her. Nina immediately runs after Marie until she catches up with her. The time, \(t \mathrm {~s}\), is measured from the moment when Nina starts running. So when \(t = 0\), both girls are at O .
Nina's acceleration, \(a \mathrm {~ms} ^ { - 2 }\), is given by $$\begin{array} { l l } a = 4 - t & \text { for } 0 \leqslant t \leqslant 4 , \\ a = 0 & \text { for } t > 4 . \end{array}$$

1. Show that Nina's speed, \(v \mathrm {~ms} ^ { - 1 }\), is given by $$\begin{array} { l l } v = 4 t - \frac { 1 } { 2 } t ^ { 2 } & \text { for } 0 \leqslant t \leqslant 4 , \\ v = 8 & \text { for } t > 4 . \end{array}$$
2. Find an expression for the distance Nina has run at time \(t\), for \(0 \leqslant t \leqslant 4\). Find how far Nina has run when \(t = 4\) and when \(t = 5 \frac { 1 } { 3 }\).
3. Show that Nina catches up with Marie when \(t = 5 \frac { 1 } { 3 }\).

8 The gradient of a curve is \(6 x ^ { 2 } + 12 x ^ { \frac { 1 } { 2 } }\). The curve passes through the point \(( 4,10 )\). Find the equation of the curve.

11 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - x ^ { 2 }\). The curve passes through the point \(( 6,1 )\). Find the equation of the curve.

2 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x + 3\). The curve passes through the point ( 2,9 ).

1. Find the equation of the tangent to the curve at the point \(( 2,9 )\).
2. Find the equation of the curve and the coordinates of its points of intersection with the \(x\)-axis. Find also the coordinates of the minimum point of this curve.
3. Find the equation of the curve after it has been stretched parallel to the \(x\)-axis with scale factor \(\frac { 1 } { 2 }\). Write down the coordinates of the minimum point of the transformed curve. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-2_1020_940_244_679} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} Fig. 11 shows a sketch of the cubic curve \(y = \mathrm { f } ( x )\). The values of \(x\) where it crosses the \(x\)-axis are - 5 , - 2 and 2 , and it crosses the \(y\)-axis at \(( 0 , - 20 )\).
4. Express \(\mathrm { f } ( x )\) in factorised form.
5. Show that the equation of the curve may be written as \(y = x ^ { 3 } + 5 x ^ { 2 } - 4 x - 20\).
6. Use calculus to show that, correct to 1 decimal place, the \(x\)-coordinate of the minimum point on the curve is 0.4 . Find also the coordinates of the maximum point on the curve, giving your answers correct to 1 decimal place.
7. State, correct to 1 decimal place, the coordinates of the maximum point on the curve \(y = \mathrm { f } ( 2 x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-3_768_1023_223_598} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} Fig. 11 shows the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\).
8. Use calculus to find \(\int _ { 1 } ^ { 3 } \left( x ^ { 3 } - 3 x ^ { 2 } - x + 3 \right) \mathrm { d } x\) and state what this represents.
9. Find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\), giving your answers in surd form. Hence state the set of values of \(x\) for which \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\) is a decreasing function.
10. Differentiate \(x ^ { 3 } - 3 x ^ { 2 } - 9 x\). Hence find the \(x\)-coordinates of the stationary points on the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 9 x\), showing which is the maximum and which the minimum.
11. Find, in exact form, the coordinates of the points at which the curve crosses the \(x\)-axis.
12. Sketch the curve. A curve has equation \(y = x ^ { 3 } - 6 x ^ { 2 } + 12\).
13. Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.
14. Find, in the form \(y = m x + c\), the equation of the normal to the curve at the point \(( 2 , - 4 )\).

1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

A particle \(P\) is moving in a straight line.
At time \(t\) seconds, the speed, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of \(P\) is given by the continuous function $$v = \begin{cases} \sqrt { 2 t + 1 } & 0 \leqslant t \leqslant k \\ \frac { 3 } { 4 } t & t > k \end{cases}$$ where \(k\) is a constant.

1. Show that \(k = 4\), explaining your method carefully.
2. Find the acceleration of \(P\) when \(t = 1.5\) At time \(t = 0 , P\) passes through the point \(O\)
3. Find the distance of \(P\) from \(O\) when \(t = 8\)

8 \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-3_588_915_954_614} The diagram shows part of each of the curves \(y = e ^ { \frac { 1 } { 5 } x }\) and \(y = \sqrt [ 3 ] { } ( 3 x + 8 )\). The curves meet, as shown in the diagram, at the point \(P\). The region \(R\), shaded in the diagram, is bounded by the two curves and by the \(y\)-axis.

1. Show by calculation that the \(x\)-coordinate of \(P\) lies between 5.2 and 5.3.
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 5 } { 3 } \ln ( 3 x + 8 )\).
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(P\) correct to 2 decimal places.
4. Use integration, and your answer to part (iii), to find an approximate value of the area of the region \(R\). \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-4_625_647_264_749} The function f is defined by \(\mathrm { f } ( x ) = \sqrt { } ( m x + 7 ) - 4\), where \(x \geqslant - \frac { 7 } { m }\) and \(m\) is a positive constant. The diagram shows the curve \(y = \mathrm { f } ( x )\).
5. A sequence of transformations maps the curve \(y = \sqrt { } x\) to the curve \(y = \mathrm { f } ( x )\). Give details of these transformations.
6. Explain how you can tell that f is a one-one function and find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
7. It is given that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) do not meet. Explain how it can be deduced that neither curve meets the line \(y = x\), and hence determine the set of possible values of \(m\). [5]