1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

A curve has equation \(y = \frac{3 \cos x}{2 + \sin x}\), for \(-\frac{1}{2}\pi \leqslant x \leqslant \frac{1}{2}\pi\).

1. Find the exact coordinates of the stationary point of the curve. [6]
2. The constant \(a\) is such that \(\int_0^a \frac{3 \cos x}{2 + \sin x} \, dx = 1\). Find the value of \(a\), giving your answer correct to 3 significant figures. [4]

\includegraphics{figure_7} The diagram shows the curve \(y = 5\sin^2 x \cos^3 x\) for \(0 \leqslant x \leqslant \frac{1}{2}\pi\), and its maximum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places. [5]
2. Using the substitution \(u = \sin x\) and showing all necessary working, find the exact area of \(R\). [4]

Let \(f(x) = \frac{6x^2 + 8x + 9}{(2 - x)(3 + 2x)^2}\).

1. Express \(f(x)\) in partial fractions. [5]
2. Hence, showing all necessary working, show that \(\int_{-1}^0 f(x) dx = 1 + \frac{1}{2}\ln\left(\frac{4}{3}\right)\). [5]

A particle \(P\) moves in a straight line and passes through a point \(O\) of the line with velocity \(2\text{ m s}^{-1}\). At time \(t\) s after passing through \(O\), the velocity of \(P\) is \(v\text{ m s}^{-1}\) and the acceleration of \(P\) is given by \(\text{e}^{-0.5t}\text{ m s}^{-2}\). Calculate the velocity of \(P\) when \(t = 1.2\). [4]

1. Show that \(\frac{(3 - \sqrt{x})^2}{\sqrt{x}}\) can be written as \(9x^{-\frac{1}{2}} - 6 + x^{\frac{1}{2}}\). [2]

Given that \(\frac{dy}{dx} = \frac{(3 - \sqrt{x})^2}{\sqrt{x}}\), \(x > 0\), and that \(y = \frac{2}{3}\) at \(x = 1\),

1. find \(y\) in terms of \(x\). [6]

The curve with equation \(y = f(x)\) passes through the point \((1, 6)\). Given that $$f'(x) = 3 + \frac{5x^2 + 2}{x^4}, \quad x > 0,$$ find \(f(x)\) and simplify your answer. [7]

\includegraphics{figure_2} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = \tan \theta, \quad y = 1 + 2\cos 2\theta, \quad 0 \leq \theta < \frac{\pi}{2}$$ The curve \(C\) crosses the \(x\)-axis at \((\sqrt{3}, 0)\). The finite shaded region \(S\) shown in Figure 2 is bounded by \(C\), the line \(x = 1\) and the \(x\)-axis. This shaded region is rotated through \(2\pi\) radians about the \(x\)-axis to form a solid of revolution.

1. Show that the volume of the solid of revolution formed is given by the integral $$k \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (16 \cos^2 \theta - 8 + \sec^2 \theta) \, d\theta$$ where \(k\) is a constant. [5]
2. Hence, use integration to find the exact value for this volume. [5]

\includegraphics{figure_2} Figure 2 shows a sketch of the curve with equation \(y = \sqrt{(3-x)(x+1)}\), \(0 \leqslant x \leqslant 3\) The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis, and the \(y\)-axis.

1. Use the substitution \(x = 1 + 2\sin\theta\) to show that $$\int_0^3 \sqrt{(3-x)(x+1)} dx = k \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2\theta d\theta$$ where \(k\) is a constant to be determined. [5]
2. Hence find, by integration, the exact area of \(R\). [3]

$$\frac{dy}{dx} = 5 + \frac{1}{x^2}.$$

1. Use integration to find \(y\) in terms of \(x\). [3]
2. Given that \(y = 7\) when \(x = 1\), find the value of \(y\) at \(x = 2\). [4]

$$\frac{dy}{dx} = 5 + \frac{1}{x^2}.$$

1. Use integration to find \(y\) in terms of \(x\). [3]
2. Given that \(y = 7\) when \(x = 1\), find the value of \(y\) at \(x = 2\). [4]

The gradient of a curve is given by \(\frac{dy}{dx} = \frac{18}{x} + 2\). The curve passes through the point \((3, 6)\). Find the equation of the curve. [5]

Show that \(\int_2^8 \frac{3}{x} \, dx = \ln 64\). [4]

Find \(\int \frac{10}{(2x - 7)^2} \, dx\). [3]

Given that $$\int_0^{\ln 4} (ke^{3x} + (k - 2)e^{-\frac{x}{3}}) \, dx = 185,$$ find the value of the constant \(k\). [7]

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]

Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = \frac{1}{\cos^2 x}\), \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), together with its asymptotes \(x = \frac{1}{2}\pi\) and \(x = -\frac{1}{2}\pi\). \includegraphics{figure_9}

1. Use the quotient rule to show that the derivative of \(\frac{\sin x}{\cos x}\) is \(\frac{1}{\cos^2 x}\). [3]
2. Find the area bounded by the curve \(y = f(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\pi\). [3]

The function \(g(x)\) is defined by \(g(x) = \frac{1}{2}f(x + \frac{1}{4}\pi)\).

1. Verify that the curves \(y = f(x)\) and \(y = g(x)\) cross at \((0, 1)\). [3]
2. State a sequence of two transformations such that the curve \(y = f(x)\) is mapped to the curve \(y = g(x)\). On the copy of Fig. 9, sketch the curve \(y = g(x)\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve. [8]
3. Use your result from part (ii) to write down the area bounded by the curve \(y = g(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = -\frac{1}{4}\pi\). [1]

Evaluate \(\int_0^{\frac{\pi}{4}} (1 - \sin 3x) \, dx\), giving your answer in exact form. [3]

Find the exact value of \(\int_0^{\frac{1}{4}\pi} (1 + \cos \frac{1}{2}x) dx\). [3]

Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = e^{2x} + k e^{-2x}\) and \(k\) is a constant greater than 1. The curve crosses the \(y\)-axis at P and has a turning point Q. \includegraphics{figure_9}

1. Find the \(y\)-coordinate of P in terms of \(k\). [1]
2. Show that the \(x\)-coordinate of Q is \(\frac{1}{4}\ln k\), and find the \(y\)-coordinate in its simplest form. [5]
3. Find, in terms of \(k\), the area of the region enclosed by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\ln k\). Give your answer in the form \(ak + b\). [4]

The function \(g(x)\) is defined by \(g(x) = f(x + \frac{1}{4}\ln k)\).

1. 1. Show that \(g(x) = \sqrt{k}(e^{2x} + e^{-2x})\). [3]
   2. Hence show that \(g(x)\) is an even function. [2]
   3. Deduce, with reasons, a geometrical property of the curve \(y = f(x)\). [3]