1.08b Integrate x^n: where n != -1 and sums

6 A curve \(C\) is defined for \(x > 0\) by the equation \(y = x + 1 + \frac { 4 } { x ^ { 2 } }\) and is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{c16d94a6-52f2-4bf3-acee-0b227ae55a1a-4_545_784_420_628}

1. 1. Given that \(y = x + 1 + \frac { 4 } { x ^ { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. The curve \(C\) has a minimum point \(M\). Find the coordinates of \(M\).
   3. Find an equation of the normal to \(C\) at the point ( 1,6 ).
2. 1. Find \(\int \left( x + 1 + \frac { 4 } { x ^ { 2 } } \right) \mathrm { d } x\).
   2. Hence find the area of the region bounded by the curve \(C\), the lines \(x = 1\) and \(x = 4\) and the \(x\)-axis.

1

1. Simplify:
   1. \(x ^ { \frac { 3 } { 2 } } \times x ^ { \frac { 1 } { 2 } }\);
   2. \(x ^ { \frac { 3 } { 2 } } \div x\);
   3. \(\left( x ^ { \frac { 3 } { 2 } } \right) ^ { 2 }\).
2. 1. Find \(\int 3 x ^ { \frac { 1 } { 2 } } \mathrm {~d} x\).
   2. Hence find the value of \(\int _ { 1 } ^ { 9 } 3 x ^ { \frac { 1 } { 2 } } \mathrm {~d} x\).

1 A particle moves along a straight line. At time \(t\), it has velocity \(v\), where $$v = 4 t ^ { 3 } - 8 \sin 2 t + 5$$ When \(t = 0\), the particle is at the origin.
Find an expression for the displacement of the particle from the origin at time \(t\).

4 A particle moves so that at time \(t\) seconds its velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) is given by $$\mathbf { v } = \left( 4 t ^ { 3 } - 12 t + 3 \right) \mathbf { i } + 5 \mathbf { j } + 8 t \mathbf { k }$$

1. When \(t = 0\), the position vector of the particle is \(( - 5 \mathbf { i } + 6 \mathbf { k } )\) metres. Find the position vector of the particle at time \(t\).
2. Find the acceleration of the particle at time \(t\).
3. Find the magnitude of the acceleration of the particle at time \(t\). Do not simplify your answer.
4. Hence find the time at which the magnitude of the acceleration is a minimum.
5. The particle is moving under the action of a single variable force \(\mathbf { F }\) newtons. The mass of the particle is 7 kg . Find the minimum magnitude of \(\mathbf { F }\).

1 A particle moves in a straight line and at time \(t\) seconds has velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where $$v = 6 t ^ { 2 } + 4 t - 7 , \quad t \geqslant 0$$

1. Find an expression for the acceleration of the particle at time \(t\).
2. The mass of the particle is 3 kg . Find the resultant force on the particle when \(t = 4\).
3. When \(t = 0\), the displacement of the particle from the origin is 5 metres. Find an expression for the displacement of the particle from the origin at time \(t\).

3

1. Amaya and Ben integrated \(( 1 + x ) ^ { 2 }\), with respect to \(x\), using different methods, as follows. Amaya: \(\quad \int ( 1 + x ) ^ { 2 } \mathrm {~d} x = \frac { ( 1 + x ) ^ { 3 } } { 3 } + c \quad = \frac { 1 } { 3 } + x + x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } + c\) Ben: \(\quad \int ( 1 + x ) ^ { 2 } \mathrm {~d} x = \int \left( 1 + 2 x + x ^ { 2 } \right) \mathrm { d } x = x + x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } + c\) Charlie said that, because these answers are different, at least one of them must be wrong. Explain whether you agree with Charlie's statement.
2. You are given that \(a\) is a constant greater than 1 .
   1. Find \(\int _ { 1 } ^ { a } \frac { 1 } { ( 1 + x ) ^ { 2 } } \mathrm {~d} x\), giving your answer as a single fraction in terms of the constant \(a\).
   2. You are given that the area enclosed by the curve \(y = \frac { 1 } { ( 1 + x ) ^ { 2 } }\), the \(x\)-axis and the lines \(x = 1\) and \(x = a\) is equal to \(\frac { 1 } { 3 }\). Determine the value of \(a\).
3. In this question you must show detailed reasoning. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 12 } \pi } \frac { \cos 2 x } { \sin 2 x + 2 } \mathrm {~d} x\), giving your answer in its simplest form.

3 It is given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { x }$$ Find an expression for \(y\).
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1 Find \(\int 12 x ^ { 3 } \mathrm {~d} x\) Circle your answer. \(36 x ^ { 2 } + c\) \(3 x ^ { 4 } + c\) \(3 x ^ { 2 } + c\) \(36 x ^ { 4 } + c\)

5 The binomial expansion of \(( 2 + 5 x ) ^ { 4 }\) is given by $$( 2 + 5 x ) ^ { 4 } = A + 160 x + B x ^ { 2 } + 1000 x ^ { 3 } + 625 x ^ { 4 }$$ 5

1. Find the value of \(A\) and the value of \(B\).
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   5
2. Show that $$( 2 + 5 x ) ^ { 4 } - ( 2 - 5 x ) ^ { 4 } = C x + D x ^ { 3 }$$ where \(C\) and \(D\) are constants to be found.
   5
3. Hence, or otherwise, find $$\int \left( ( 2 + 5 x ) ^ { 4 } - ( 2 - 5 x ) ^ { 4 } \right) \mathrm { d } x$$

1. In this question you must show all stages of your working.

$$f ( x ) = \frac { ( x + 5 ) ^ { 2 } } { \sqrt { x } } \quad x > 0$$

1. Find \(\int f ( x ) d x\)
2. 1. Show that when \(\mathrm { f } ^ { \prime } ( x ) = 0\) $$3 x ^ { 2 } + 10 x - 25 = 0$$
   2. Hence state the value of \(x\) for which $$\mathrm { f } ^ { \prime } ( x ) = 0$$

5

1. Find \(\int \left( 3 x ^ { 2 } - 4 x + 8 \right) \mathrm { d } x\).
2. Hence find \(\int _ { 1 } ^ { 3 } \left( 3 x ^ { 2 } - 4 x + 8 \right) \mathrm { d } x\).

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]

The equation of a curve is such that \(\frac{dy}{dx} = 4x - 3\sqrt{x} + 1\).

1. Find the \(x\)-coordinate of the point on the curve at which the gradient is \(\frac{11}{2}\). [3]
2. Given that the curve passes through the point \((4, 11)\), find the equation of the curve. [4]

\includegraphics{figure_7} The diagram shows part of the curve with equation \(y = \frac{12}{\sqrt{2x+1}}\). The point \(A\) on the curve has coordinates \(\left(\frac{7}{2}, 6\right)\).

1. Find the equation of the tangent to the curve at \(A\). Give your answer in the form \(y = mx + c\). [4]
2. Find the area of the region bounded by the curve and the lines \(x = 0\), \(x = \frac{7}{2}\) and \(y = 0\). [4]

A function f with domain \(x > 0\) is such that \(\mathrm{f}'(x) = 8(2x - 3)^{\frac{1}{3}} - 10x^{\frac{3}{5}}\). It is given that the curve with equation \(y = \mathrm{f}(x)\) passes through the point \((1, 0)\).

1. Find the equation of the normal to the curve at the point \((1, 0)\). [3]
2. Find f\((x)\). [4]

It is given that the equation \(\mathrm{f}'(x) = 0\) can be expressed in the form $$125x^2 - 128x + 192 = 0.$$

1. Determine, making your reasoning clear, whether f is an increasing function, a decreasing function or neither. [3]

Find \(\int \left(x^3 + \frac{1}{x^3}\right) \mathrm{d}x\). [3]

\includegraphics{figure_9} The diagram shows part of the curve \(y = -x^2 + 8x - 10\) which passes through the points \(A\) and \(B\). The curve has a maximum point at \(A\) and the gradient of the line \(BA\) is \(2\).

1. Find the coordinates of \(A\) and \(B\). [7]
2. Find \(\int y \, dx\) and hence evaluate the area of the shaded region. [4]

A curve is such that \(\frac{d^2y}{dx^2} = -4x\). The curve has a maximum point at \((2, 12)\).

1. Find the equation of the curve. [6]

A point \(P\) moves along the curve in such a way that the \(x\)-coordinate is increasing at 0.05 units per second.

1. Find the rate at which the \(y\)-coordinate is changing when \(x = 3\), stating whether the \(y\)-coordinate is increasing or decreasing. [2]

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]

\includegraphics{figure_11} The diagram shows part of the curve \(y = 3\sqrt{(4x + 1)} - 2x\). The curve crosses the \(y\)-axis at \(A\) and the stationary point on the curve is \(M\).

1. Obtain expressions for \(\frac{\text{d}y}{\text{d}x}\) and \(\int y \text{d}x\). [5]
2. Find the coordinates of \(M\). [3]
3. Find, showing all necessary working, the area of the shaded region. [4]

A curve is such that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{(4x + 1)}\) and \((2, 5)\) is a point on the curve.

1. Find the equation of the curve. [4]
2. A point \(P\) moves along the curve in such a way that the \(y\)-coordinate is increasing at a constant rate of 0.06 units per second. Find the rate of change of the \(x\)-coordinate when \(P\) passes through \((2, 5)\). [2]
3. Show that \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \times \frac{\mathrm{d}y}{\mathrm{d}x}\) is constant. [2]

A curve with equation \(y = \mathrm{f}(x)\) passes through the point \(A(3, 1)\) and crosses the \(y\)-axis at \(B\). It is given that \(\mathrm{f}'(x) = (3x - 1)^{-\frac{1}{3}}\). Find the \(y\)-coordinate of \(B\). [6]

A particle moves in a straight line starting from a point \(O\) with initial velocity \(1\) m s\(^{-1}\). The acceleration of the particle at time \(t\) s after leaving \(O\) is \(a\) m s\(^{-2}\), where $$a = 1.2t^{\frac{1}{2}} - 0.6t.$$

1. At time \(T\) s after leaving \(O\) the particle reaches its maximum velocity. Find the value of \(T\). [2]
2. Find the velocity of the particle when its acceleration is maximum (you do not need to verify that the acceleration is a maximum rather than a minimum). [6]