1.08d Evaluate definite integrals: between limits

\includegraphics{figure_9} The diagram shows the curve with equation \(y = 2 \ln(x - 1)\). The point \(P\) has coordinates \((0, p)\). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 0\), \(y = 0\) and \(y = p\). The units on the axes are centimetres. The region \(R\) is rotated completely about the \(y\)-axis to form a solid.

1. Show that the volume, \(V \text{ cm}^3\), of the solid is given by $$V = \pi(e^p + 4e^{\frac{p}{2}} + p - 5).$$ [8]
2. It is given that the point \(P\) is moving in the positive direction along the \(y\)-axis at a constant rate of \(0.2 \text{ cm min}^{-1}\). Find the rate at which the volume of the solid is increasing at the instant when \(p = 4\), giving your answer correct to 2 significant figures. [5]

\includegraphics{figure_6} The diagram shows the curve with equation \(y = \frac{1}{\sqrt{3x + 2}}\). The shaded region is bounded by the curve and the lines \(x = 0\), \(x = 2\) and \(y = 0\).

1. Find the exact area of the shaded region. [4]
2. The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed, simplifying your answer. [5]

It is given that \(\int_a^{3a} (e^{5x} + e^x) dx = 100\), where \(a\) is a positive constant.

1. Show that \(a = \frac{1}{5}\ln(300 + 3e^a - 2e^{3a})\). [5]
2. Use an iterative process, based on the equation in part (i), to find the value of \(a\) correct to 4 decimal places. Use a starting value of 0.6 and show the result of each step of the process. [4]

\includegraphics{figure_4} The diagram shows part of the curve \(y = \frac{k}{x}\), where \(k\) is a positive constant. The points A and B on the curve have \(x\)-coordinates 2 and 6 respectively. Lines through A and B parallel to the axes as shown meet at the point C. The region R is bounded by the curve and the lines \(x = 2\), \(x = 6\) and \(y = 0\). The region S is bounded by the curve and the lines AC and BC. It is given that the area of the region R is \(\ln 81\).

1. Show that \(k = 4\). [3]
2. Find the exact volume of the solid produced when the region S is rotated completely about the \(x\)-axis. [4]

Fig. 9 shows the curve \(y = f(x)\). The endpoints of the curve are P \((-\pi, 1)\) and Q \((\pi, 3)\), and \(f(x) = a + \sin bx\), where \(a\) and \(b\) are constants. \includegraphics{figure_9}

1. Using Fig. 9, show that \(a = 2\) and \(b = \frac{1}{2}\). [3]
2. Find the gradient of the curve \(y = f(x)\) at the point \((0, 2)\). Show that there is no point on the curve at which the gradient is greater than this. [5]
3. Find \(f^{-1}(x)\), and state its domain and range. Write down the gradient of \(y = f^{-1}(x)\) at the point \((2, 0)\). [6]
4. Find the area enclosed by the curve \(y = f(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\). [4]

\includegraphics{figure_7} The diagram shows the curve with equation \(y = 2x - e^{\frac{1}{2}x}\). The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 4\).

1. Find the area of the shaded region, giving your answer in terms of e. [4]

The shaded region is rotated through four right angles about the \(x\)-axis.

1. Using Simpson's rule with two strips, estimate the volume of the solid formed. [5]

Show that $$\int_1^7 \frac{2}{4x-1} \, dx = \ln 3.$$ [4]

Fig. 9 shows the curve \(y = \frac{x^2}{3x - 1}\). P is a turning point, and the curve has a vertical asymptote \(x = a\). \includegraphics{figure_1}

1. Write down the value of \(a\). [1]
2. Show that \(\frac{dy}{dx} = \frac{x(3x - 2)}{(3x - 1)^2}\) [3]
3. Find the exact coordinates of the turning point P. Calculate the gradient of the curve when \(x = 0.6\) and \(x = 0.8\), and hence verify that P is a minimum point. [7]
4. Using the substitution \(u = 3x - 1\), show that \(\int \frac{x^2}{3x - 1} dx = \frac{1}{27} \int \left( u + 2 + \frac{1}{u} \right) du\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac{2}{3}\) and \(x = 1\). [7]

Fig. 8 shows part of the curve \(y = \text{f}(x)\), where \(\text{f}(x) = e^{-\frac{1}{5}x} \sin x\), for all \(x\). \includegraphics{figure_8}

1. Sketch the graphs of (A) \(y = \text{f}(2x)\), (B) \(y = \text{f}(x + \pi)\). [4]
2. Show that the \(x\)-coordinate of the turning point P satisfies the equation \(\tan x = 5\). Hence find the coordinates of P. [6]
3. Show that \(\text{f}(x + \pi) = -e^{-\frac{1}{5}\pi}\text{f}(x)\). Hence, using the substitution \(u = x - \pi\), show that $$\int_{\pi}^{2\pi} \text{f}(x)\,dx = -e^{-\frac{1}{5}\pi} \int_{0}^{\pi} \text{f}(u)\,du.$$ Interpret this result graphically. [You should not attempt to integrate f(x).] [8]

Show that \(\int_1^4 \frac{x}{x^2 + 2} \, dx = \frac{1}{2} \ln 6\). [4]

\includegraphics{figure_2} Figure 2 shows a sketch of the curve \(C\) with equation \(y = \frac{4}{x - 3}\), \(x \neq 3\). The points \(A\) and \(B\) on the curve have \(x\)-coordinates 3.25 and 5 respectively.

1. Write down the \(y\)-coordinates of \(A\) and \(B\). [1]
2. Show that an equation of \(C\) is \(\frac{3y + 4}{y} = 0\), \(y \neq 0\). [1]

The shaded region \(R\) is bounded by \(C\), the \(y\)-axis and the lines through \(A\) and \(B\) parallel to the \(x\)-axis. The region \(R\) is rotated through 360° about the \(y\)-axis to form a solid shape \(S\).

1. Find the volume of \(S\), giving your answer in the form \(\pi(a + b \ln c)\), where \(a\), \(b\) and \(c\) are integers. [7]

The solid shape \(S\) is used to model a cooling tower. Given that 1 unit on each axis represents 3 metres,

1. show that the volume of the tower is approximately 15500 m\(^3\). [2]

\includegraphics{figure_1} In Fig. 1, the curve \(C\) has equation \(y = f(x)\), where $$f(x) = x + \frac{2}{x^2}, \quad x > 0.$$ The shaded region is bounded by \(C\), the \(x\)-axis and the lines with equations \(x = 1\) and \(x = 2\). The shaded region is rotated through \(2\pi\) radians about the \(x\)-axis. Using calculus, calculate the volume of the solid generated. Give your answer in the form \(\pi(a + \ln b)\), where \(a\) and \(b\) are constants. [8]

\includegraphics{figure_1} The curve \(C\) has equation \(y = f(x)\), \(x \in \mathbb{R}\). Figure 1 shows the part of \(C\) for which \(0 \leq x \leq 2\). Given that $$\frac{dy}{dx} = e^x - 2x^2,$$ and that \(C\) has a single maximum, at \(x = k\),

1. show that \(1.48 < k < 1.49\). [3]

Given also that the point \((0, 5)\) lies on \(C\),

1. find \(f(x)\). [4]

The finite region \(R\) is bounded by \(C\), the coordinate axes and the line \(x = 2\).

1. Use integration to find the exact area of \(R\). [4]

A student tests the accuracy of the trapezium rule by evaluating \(I\), where $$I = \int_{0.5}^{1.5} \left(\frac{3}{x} + x^4\right) dx.$$

1. Complete the student's table, giving values to 2 decimal places where appropriate.

   \(x\)	0.5	0.75	1	1.25	1.5
   \(\frac{3}{x} + x^4\)	6.06	4.32

   [2]
2. Use the trapezium rule, with all the values from your table, to calculate an estimate for the value of \(I\). [4]
3. Use integration to calculate the exact value of \(I\). [4]
4. Verify that the answer obtained by the trapezium rule is within 3\% of the exact value. [2]

\includegraphics{figure_1} Figure 1 shows a cross-section \(R\) of a dam. The line \(AC\) is the vertical face of the dam, \(AB\) is the horizontal base and the curve \(BC\) is the profile. Taking \(x\) and \(y\) to be the horizontal and vertical axes, then \(A\), \(B\) and \(C\) have coordinates \((0, 0)\), \((3\pi^2, 0)\) and \((0, 30)\) respectively. The area of the cross-section is to be calculated. Initially the profile \(BC\) is approximated by a straight line.

1. Find an estimate for the area of the cross-section \(R\) using this approximation. [1]

The profile \(BC\) is actually described by the parametric equations. $$x = 16t^2 - \pi^2, \quad y = 30 \sin 2t, \quad \frac{\pi}{4} \leq t \leq \frac{\pi}{2}.$$

1. Find the exact area of the cross-section \(R\). [7]
2. Calculate the percentage error in the estimate of the area of the cross-section \(R\) that you found in part (a). [2]

Evaluate $$\int_0^{\frac{\pi}{4}} \sin 2x \cos x \, dx.$$ [5]

A motorcyclist starts from rest at a point \(O\) and travels in a straight line. His velocity after \(t\) seconds is \(v\) m s\(^{-1}\), for \(0 \leq t \leq T\), where \(v = 7.2t - 0.45t^2\). The motorcyclist's acceleration is zero when \(t = T\).

1. Find the value of \(T\). [4]
2. Show that \(v = 28.8\) when \(t = T\). [1]

For \(t \geq T\) the motorcyclist travels in the same direction as before, but with constant speed \(28.8\) m s\(^{-1}\).

1. Find the displacement of the motorcyclist from \(O\) when \(t = 31\). [6]

A cyclist travels along a straight road. Her velocity \(v\) m s\(^{-1}\), at time \(t\) seconds after starting from a point \(O\), is given by \(v = 2\) for \(0 \leq t \leq 10\), \(v = 0.03t^2 - 0.3t + 2\) for \(t \geq 10\).

1. Find the displacement of the cyclist from \(O\) when \(t = 10\). [1]
2. Show that, for \(t \geq 10\), the displacement of the cyclist from \(O\) is given by the expression \(0.01t^3 - 0.15t^2 + 2t + 5\). [4]
3. Find the time when the acceleration of the cyclist is \(0.6\) m s\(^{-2}\). Hence find the displacement of the cyclist from \(O\) when her acceleration is \(0.6\) m s\(^{-2}\). [5]

\includegraphics{figure_7} A sprinter \(S\) starts from rest at time \(t = 0\), where \(t\) is in seconds, and runs in a straight line. For \(0 \leq t \leq 3\), \(S\) has velocity \((6t - t^2)\) m s\(^{-1}\). For \(3 < t \leq 22\), \(S\) runs at a constant speed of \(9\) m s\(^{-1}\). For \(t > 22\), \(S\) decelerates at \(0.6\) m s\(^{-2}\) (see diagram).

1. Express the acceleration of \(S\) during the first \(3\) seconds in terms of \(t\). [2]
2. Show that \(S\) runs \(18\) m in the first \(3\) seconds of motion. [5]
3. Calculate the time \(S\) takes to run \(100\) m. [3]
4. Calculate the time \(S\) takes to run \(200\) m. [7]

The velocity, \(v\) ms\(^{-1}\), of a particle at time \(t\) s is given by \(v = 4t^2 - 9\).

1. Find the acceleration of the particle when it is instantaneously at rest. [3 marks]
2. Find the distance travelled by the particle from time \(t = 0\) until it comes to rest. [4 marks]

A particle \(P\) of mass \(0.5\) kg falls vertically from rest. After \(t\) seconds it has speed \(v\) m s\(^{-1}\). A resisting force of magnitude \(1.5v\) newtons acts on \(P\) as it falls.

1. Show that \(3v = 9.8(1 - e^{-3t})\). [8]
2. Find the distance that \(P\) falls in the first two seconds of its motion. [5]