1.08i - OCR Spec

OCR Further Additional Pure 2017 Specimen Q5

9 marks Challenging +1.2

In this question you must show detailed reasoning. It is given that $I_n = \int_0^\pi \sin^n \theta \, d\theta$ for $n \geq 0$.

Prove that $I_n = \frac{n-1}{n} I_{n-2}$ for $n \geq 2$. [5]
Evaluate $I_1$ and use the reduction formula to determine the exact value of $\int_0^\pi \cos^2 \theta \sin^5 \theta \, d\theta$. [4]

Pre-U Pre-U 9794/2 2010 June Q9

15 marks Challenging +1.2

Show that $$\int x^n \ln x \, dx = \frac{x^{n+1}}{(n+1)^2}\left((n+1)\ln a - 1\right) + \frac{1}{(n+1)^2},$$ where $n \neq -1$ and $a > 1$. [6]
1. Determine the $x$-coordinate of the point of intersection of the curves $y = x^3 \ln x$ and $y = x \ln 2^x$, where $x > 0$. [2]
2. Find the exact value of the area of the region enclosed between these two curves, the line $x = 1$ and their point of intersection. Express your answer in the form $b + c \ln 2$, where $b$ and $c$ are rational. [4]
The curve $y = (x^3 \ln x)^{0.5}$, for $1 < x < e$, is rotated through $2\pi$ radians about the $x$-axis. Determine the value of the resulting volume of revolution, giving your answer correct to 4 significant figures. [3]

Pre-U Pre-U 9794/2 2011 June Q3

5 marks Moderate -0.8

Use integration by parts to find $\int x \sin 3x \, dx$. [5]

Edexcel AEA 2014 June Q6

20 marks Hard +2.3

A curve with equation $y = f(x)$ has $f(x) \geq 0$ for $x \geq a$ and $$A = \int_a^b f(x) \, dx \quad \text{and} \quad V = \pi \int_a^b [f(x)]^2 \, dx$$ where $a$ and $b$ are constants with $b > a$. Use integration by substitution to show that for the positive constants $r$ and $h$ $$\pi \int_{a+h}^{b+h} [r + f(x - h)]^2 \, dx = \pi r^2 (b - a) + 2\pi rA + V$$ [3]
% \includegraphics{figure_1} - Shows a curve with vertical asymptotes at x=m and x=n, crossing y-axis at point p Figure 1 shows part of the curve $C$ with equation $y = 4 + \frac{2}{\sqrt{3}\cos x + \sin x}$ This curve has asymptotes $x = m$ and $x = n$ and crosses the $y$-axis at $(0, p)$. (a) Find the value of $p$, the value of $m$ and the value of $n$. [4] (b) Show that the equation of $C$ can be written in the form $y = r + f(x - h)$ and specify the function $f$ and the constants $r$ and $h$. [4] The region bounded by $C$, the $x$-axis and the lines $x = \frac{\pi}{6}$ and $x = \frac{\pi}{3}$ is rotated through $2\pi$ radians about the $x$-axis. (c) Find the volume of the solid formed. [9]

Edexcel AEA 2014 June Q7

23 marks Hard +2.3

% \includegraphics{figure_2} - Shows a circular tower with center T at (0,1), a goat at point G attached to the base at O, with string along arc OA then tangent AG A circular tower stands in a large horizontal field of grass. A goat is attached to one end of a string and the other end of the string is attached to the fixed point $O$ at the base of the tower. Taking the point $O$ as the origin $(0, 0)$, the centre of the base of the tower is at the point $T(0, 1)$. The radius of the base of the tower is 1. The string has length $\pi$ and you may ignore the size of the goat. The curve $C$ represents the edge of the region that the goat can reach as shown in Figure 2.

Write down the equation of $C$ for $y < 0$. [1] When the goat is at the point $G(x, y)$, with $x > 0$ and $y > 0$, as shown in Figure 2, the string lies along $OAG$ where $OA$ is an arc of the circle with angle $OTA = \theta$ radians and $AG$ is a tangent to the circle at $A$.
With the aid of a suitable diagram show that $$x = \sin \theta + (\pi - \theta) \cos \theta$$ $$y = 1 - \cos \theta + (\pi - \theta) \sin \theta$$ [5]
By considering $\int y \frac{dx}{d\theta} d\theta$, show that the area between $C$, the positive $x$-axis and the positive $y$-axis can be expressed in the form $$\int_0^{\pi} u \sin u \, du + \int_0^{\pi} u^2 \sin^2 u \, du + \int_0^{\pi} u \sin u \cos u \, du$$ [5]
Show that $\int_0^{\pi} u^2 \sin^2 u \, du = \frac{\pi^3}{6} + \int_0^{\pi} u \sin u \cos u \, du$ [4]
Hence find the area of grass that can be reached by the goat. [8]

Edexcel AEA 2015 June Q7

19 marks Hard +2.3

Use the substitution $x = \sec\theta$ to show that $$\int_{\sqrt{2}}^{2} \frac{1}{(x^2 - 1)^{\frac{3}{2}}} \, dx = \frac{\sqrt{6} - 2}{\sqrt{3}}$$ [5]
Use integration by parts to show that $$\int \cos\theta \cot^2\theta \, d\theta = \frac{1}{2}[\ln|\cos\theta + \cot\theta| - \cos\theta \cot\theta] + c$$ [6] % Figure shows a curve y = 1/(x^2-1)^(1/2) for x > 1, with shaded region R between x = sqrt(2) and x = 2 \includegraphics{figure_2} Figure 2 shows a sketch of part of the curve with equation $y = \frac{1}{(x^2 - 1)^{\frac{1}{2}}}$ for $x > 1$ The region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis and the lines $x = \sqrt{2}$ and $x = 2$ The region $R$ is rotated through $2\pi$ radians about the $x$-axis.
Show that the volume of the solid formed is $$\pi \left[\frac{3}{8}\ln\left(\frac{1 + \sqrt{2}}{\sqrt{3}}\right) + \frac{7}{36} - \frac{\sqrt{2}}{8}\right]$$ [8]

1.08i Integration by parts