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

6. (a) Use the substitution \(u = \sqrt { t }\) to show that $$\int _ { 1 } ^ { x } \frac { \ln t } { \sqrt { t } } \mathrm {~d} t = 4 - 4 \sqrt { x } + 2 \sqrt { x } \ln x \quad x \geqslant 1$$ (b) The function g is such that $$\int _ { 1 } ^ { x } \mathrm {~g} ( t ) \mathrm { d } t = x - \sqrt { x } \ln x - 1 \quad x \geqslant 1$$

1. Use differentiation to find the function g .
2. Evaluate \(\int _ { 4 } ^ { 16 } \mathrm {~g} ( t ) \mathrm { d } t\) and simplify your answer.
   (c) Find the value of \(x\) (where \(x > 1\) ) that gives the maximum value of $$\int _ { x } ^ { x + 1 } \frac { \ln t } { 2 ^ { t } } \mathrm {~d} t$$

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

The point \(P\) with \(x\) coordinate 3 lies on \(C\) \section*{Given}

• \(\mathrm { f } ^ { \prime } ( x ) = 4 x ^ { 2 } + k x + 3\) where \(k\) is a constant
• the normal to \(C\) at \(P\) has equation \(y = - \frac { 1 } { 24 } x + 5\)
  1. show that \(k = - 5\)
  2. Hence find \(\mathrm { f } ( x )\).

9. \begin{figure}[h] \end{figure} Figure 5 shows a sketch of part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = \frac { 6 x ^ { 2 } + 4 x - 2 } { 2 x + 1 } \quad x > - \frac { 1 } { 2 }$$

1. Find \(\mathrm { f } ^ { \prime } ( x )\), giving the answer in simplest form. The line \(l\) is the normal to \(C\) at the point \(P ( 2,6 )\)
2. Show that an equation for \(l\) is $$16 y + 5 x = 106$$
3. Write \(\mathrm { f } ( x )\) in the form \(A x + B + \frac { D } { 2 x + 1 }\) where \(A , B\) and \(D\) are constants. The region \(R\), shown shaded in Figure 5, is bounded by \(C , l\) and the \(x\)-axis.
4. Use algebraic integration to find the exact area of \(R\), giving your answer in the form \(P + Q \ln 3\), where \(P\) and \(Q\) are rational constants.
   (Solutions based entirely on calculator technology are not acceptable.)

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

The equation of a curve is such that \(\frac{dy}{dx} = \frac{4}{(x-3)^2}\) for \(x > 3\). The curve passes through the point \((4, 5)\). Find the equation of the curve. [3]

A function f is such that \(f'(x) = 6(2x-3)^2 - 6x\) for \(x \in \mathbb{R}\).

1. Determine the set of values of \(x\) for which f\((x)\) is decreasing. [4]
2. Given that f\((1) = -1\), find f\((x)\). [4]

The equation of a curve is such that \(\frac{dy}{dx} = \frac{1}{2}x + \frac{72}{x^4}\). The curve passes through the point \(P(2, 8)\).

1. Find the equation of the normal to the curve at \(P\). [2]
2. Find the equation of the curve. [4]

The function f is such that \(\mathrm{f}'(x) = 5 - 2x^2\) and \((3, 5)\) is a point on the curve \(y = \mathrm{f}(x)\). Find \(\mathrm{f}(x)\). [3]

The function f is defined for \(x \geqslant 0\). It is given that f has a minimum value when \(x = 2\) and that \(\text{f}''(x) = (4x + 1)^{-\frac{1}{2}}\).

1. Find \(\text{f}'(x)\). [3]

It is now given that \(\text{f}''(0)\), \(\text{f}'(0)\) and \(\text{f}(0)\) are the first three terms respectively of an arithmetic progression.

1. Find the value of \(\text{f}(0)\). [3]
2. Find \(\text{f}(x)\), and hence find the minimum value of f. [5]

A curve is such that \(\frac{\text{d}y}{\text{d}x} = 3x^2 + ax + b\). The curve has stationary points at \((-1, 2)\) and \((3, k)\). Find the values of the constants \(a\), \(b\) and \(k\). [8]

A curve with equation \(y = f(x)\) passes through the points \((0, 2)\) and \((3, -1)\). It is given that \(f'(x) = kx^2 - 2x\), where \(k\) is a constant. Find the value of \(k\). [5]

A curve \(y = \mathrm{f}(x)\) has a stationary point at \(P(3, -10)\). It is given that \(\mathrm{f}'(x) = 2x^2 + kx - 12\), where \(k\) is a constant.

1. Show that \(k = -2\) and hence find the \(x\)-coordinate of the other stationary point, \(Q\). [4]
2. Find \(\mathrm{f}''(x)\) and determine the nature of each of the stationary points \(P\) and \(Q\). [2]
3. Find \(\mathrm{f}(x)\). [4]

A curve is such that \(\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4\). The curve has a stationary point at \(P\) where \(x = 2\).

1. State, with a reason, the nature of this stationary point. [1]
2. Find an expression for \(\frac{dy}{dx}\). [4]
3. Given that the curve passes through the point \((1, 13)\), find the coordinates of the stationary point \(P\). [4]

A curve \(y = f(x)\) has a stationary point at \((3, 7)\) and is such that \(f''(x) = 36x^{-3}\).

1. State, with a reason, whether this stationary point is a maximum or a minimum. [1]
2. Find \(f'(x)\) and \(f(x)\). [7]

A curve is such that \(\frac{dy}{dx} = \frac{2}{a}x^{-\frac{1}{2}} + ax^{-\frac{3}{2}}\), where \(a\) is a positive constant. The point \(A(a^2, 3)\) lies on the curve. Find, in terms of \(a\),

1. the equation of the tangent to the curve at \(A\), simplifying your answer, [3]
2. the equation of the curve. [4]

It is now given that \(B(16, 8)\) also lies on the curve.

1. Find the value of \(a\) and, using this value, find the distance \(AB\). [5]

A particle \(P\) travels in a straight line, starting at rest from a point \(O\). The acceleration of \(P\) at time \(t\) s after leaving \(O\) is denoted by \(a \text{ m s}^{-2}\), where $$a = 0.3t^{\frac{1}{2}} \quad \text{for } 0 \leqslant t \leqslant 4,$$ $$a = -kt^{-\frac{1}{2}} \quad \text{for } 4 < t \leqslant T,$$ where \(k\) and \(T\) are constants.

1. Find the velocity of \(P\) at \(t = 4\). [2]

1. It is given that there is no change in the velocity of \(P\) at \(t = 4\) and that the velocity of \(P\) at \(t = 16\) is \(0.3 \text{ m s}^{-1}\). Show that \(k = 2.6\) and find an expression, in terms of \(t\), for the velocity of \(P\) for \(4 \leqslant t \leqslant T\). [4]

1. Given that \(P\) comes to instantaneous rest at \(t = T\), find the exact value of \(T\). [2]

1. Find the total distance travelled between \(t = 0\) and \(t = T\). [4]

A particle \(X\) travels in a straight line. The velocity of \(X\) at time \(t\) s after leaving a fixed point \(O\) is denoted by \(v\) m s\(^{-1}\), where $$v = -0.1t^3 + 1.8t^2 - 6t + 5.6.$$ The acceleration of \(X\) is zero at \(t = p\) and \(t = q\), where \(p < q\).

1. Find the value of \(p\) and the value of \(q\). [4]

It is given that the velocity of \(X\) is zero at \(t = 14\).

1. Find the velocities of \(X\) at \(t = p\) and at \(t = q\), and hence sketch the velocity-time graph for the motion of \(X\) for \(0 \leq t \leq 15\). [3]
2. Find the total distance travelled by \(X\) between \(t = 0\) and \(t = 15\). [5]

A particle \(P\) moves in a straight line, passing through a point \(O\) with velocity \(4.2 \text{ ms}^{-1}\). At time \(t\) s after \(P\) passes \(O\), the acceleration, \(a \text{ ms}^{-2}\), of \(P\) is given by \(a = 0.6t - 2.7\). Find the distance \(P\) travels between the times at which it is at instantaneous rest. [7]

A particle moves in a straight line. It starts from rest, at time \(t = 0\), and accelerates at 0.6 t ms\(^{-2}\) for 4 s, reaching a speed of \(V\) ms\(^{-1}\). The particle then travels at \(V\) ms\(^{-1}\) for 11 s, and finally slows down, with constant deceleration, stopping after a further 5 s.

1. Show that \(V = 4.8\). [1]
2. Sketch a velocity-time graph for the motion. [3]
3. Find an expression, in terms of \(t\), for the velocity of the particle for \(15 \leqslant t \leqslant 20\). [2]
4. Find the total distance travelled by the particle. [4]

A particle \(P\) of mass \(0.5\) kg is at rest at a point \(O\) on a rough horizontal surface. At time \(t = 0\), where \(t\) is in seconds, a horizontal force acting in a fixed direction is applied to \(P\). At time \(t\) s the magnitude of the force is \(0.6t^2\) N and the velocity of \(P\) away from \(O\) is \(v\,\text{m}\,\text{s}^{-1}\). It is given that \(P\) remains at rest at \(O\) until \(t = 0.5\).

1. Calculate the coefficient of friction between \(P\) and the surface, and show that $$\frac{\text{d}v}{\text{d}t} = 1.2t^2 - 0.3 \quad \text{for } t > 0.5.$$ [3]
2. Express \(v\) in terms of \(t\) for \(t > 0.5\). [3]
3. Find the displacement of \(P\) from \(O\) when \(t = 1.2\). [3]

A particle \(P\) of mass \(0.5\) kg is at rest at a point \(O\) on a rough horizontal surface. At time \(t = 0\), where \(t\) is in seconds, a horizontal force acting in a fixed direction is applied to \(P\). At time \(t\) s the magnitude of the force is \(0.6t^2\) N and the velocity of \(P\) away from \(O\) is \(v \text{ ms}^{-1}\). It is given that \(P\) remains at rest at \(O\) until \(t = 0.5\).

1. Calculate the coefficient of friction between \(P\) and the surface, and show that $$\frac{\text{dv}}{\text{dt}} = 1.2t^2 - 0.3 \quad \text{for } t > 0.5.$$ [3]
2. Express \(v\) in terms of \(t\) for \(t > 0.5\). [3]
3. Find the displacement of \(P\) from \(O\) when \(t = 1.2\). [3]

A particle \(P\) of mass \(0.4 \text{ kg}\) is projected horizontally along a smooth horizontal plane from a point \(O\). At time \(t \text{ s}\) after projection the velocity of \(P\) is \(v \text{ ms}^{-1}\). A force of magnitude \(0.8t \text{ N}\) directed away from \(O\) acts on \(P\) and a force of magnitude \(2e^{-t} \text{ N}\) opposes the motion of \(P\).

1. Show that \(\frac{dv}{dt} = 2t - 5e^{-t}\). [2]
2. Given that \(v = 8\) when \(t = 1\), express \(v\) in terms of \(t\). [3]
3. Find the speed of projection of \(P\). [2]

A particle moves in a straight line. At time \(t\) seconds its velocity is \(v\) ms\(^{-1}\) and its acceleration is \(a\) ms\(^{-2}\).

1. Given that \(a = —\), express \(v\) in terms of \(t\).
2. Given that \(v = tv\) when \(t = 0\), find \(v\) in terms of \(t\).
3. Find the displacement from the starting point when \(t = v\).

[6]

A curve with equation \(y = \text{f}(x)\) passes through the point \((4, 25)\) Given that $$\text{f}'(x) = \frac{3}{8}x^2 - 10x^{-\frac{1}{2}} + 1, \quad x > 0$$ find \(\text{f}(x)\), simplifying each term. [5]

The gradient of the curve \(C\) is given by $$\frac{dy}{dx} = (3x - 1)^2.$$ The point \(P(1, 4)\) lies on \(C\).

1. Find an equation of the normal to \(C\) at \(P\). [4]
2. Find an equation for the curve \(C\) in the form \(y = f(x)\). [5]
3. Using \(\frac{dy}{dx} = (3x - 1)^2\), show that there is no point on \(C\) at which the tangent is parallel to the line \(y = 1 - 2x\). [2]