1.08f Area between two curves: using integration

\includegraphics{figure_2} Figure 2 shows part of the curve C with equation y = f(x), where $$f(x) = x^3 - 6x^2 + 5x.$$ The curve crosses the x-axis at the origin O and at the points A and B.

1. Factorise f(x) completely [3 marks]
2. Write down the x-coordinates of the points A and B. [1 marks]
3. Find the gradient of C at A. [3 marks] The region R is bounded by C and the line OA, and the region S is bounded by C and the line AB.
4. Use integration to find the area of the combined regions R and S, shown shaded in Fig. 2. [7 marks]

\includegraphics{figure_2} Figure 2 shows the line with equation \(y = 9 - x\) and the curve with equation \(y = x^2 - 2x + 3\). The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). [5]

The shaded region \(R\) is bounded by the line and the curve.

1. Calculate the area of \(R\). [7]

\includegraphics{figure_2} Figure 2 shows the line with equation \(y = x + 1\) and the curve with equation \(y = 6x - x^2 - 3\). The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). [5]

The shaded region \(R\) is bounded by the line and the curve.

1. Calculate the area of \(R\). [7]

\includegraphics{figure_1} Fig. 1 shows the curve with equation \(y = 5 + 2x - x^2\) and the line with equation \(y = 2\). The curve and the line intersect at the points \(A\) and \(B\).

1. Find the \(x\)-coordinates of \(A\) and \(B\). [3]

The shaded region \(R\) is bounded by the curve and the line.

1. Find the area of \(R\). [6]

\includegraphics{figure_2} Fig. 2 shows the line with equation \(y = x + 1\) and the curve with equation \(y = 6x - x^2 - 3\). The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). [5]

The shaded region \(R\) is bounded by the line and the curve.

1. Calculate the area of \(R\). [7]

\includegraphics{figure_3} Fig. 3 shows the line with equation \(y = 9 - x\) and the curve with equation \(y = x^2 - 2x + 3\). The line and the curve intersect at the points \(A\) and \(B\), and \(O\) is the origin.

1. Calculate the coordinates of \(A\) and the coordinates of \(B\). [5]

The shaded region \(R\) is bounded by the line and the curve.

1. Calculate the area of \(R\). [7]

\includegraphics{figure_7} The diagram shows the curves \(y = -3x^2 - 9x + 30\) and \(y = x^2 + 3x - 10\).

1. Verify that the curves intersect at the points \(A(-5, 0)\) and \(B(2, 0)\). [2]
2. Show that the area of the shaded region between the curves is given by \(\int_{-5}^{2} (-4x^2 - 12x + 40) dx\). [2]
3. Hence or otherwise show that the area of the shaded region between the curves is \(228\frac{2}{3}\). [5]

Fig. 8 shows part of the curve \(y = x \sin 3x\). It crosses the \(x\)-axis at P. The point on the curve with \(x\)-coordinate \(\frac{1}{6}\pi\) is Q. \includegraphics{figure_8}

1. Find the \(x\)-coordinate of P. [3]
2. Show that Q lies on the line \(y = x\). [1]
3. Differentiate \(x \sin 3x\). Hence prove that the line \(y = x\) touches the curve at Q. [6]
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac{1}{72}(\pi^2 - 8)\). [7]

Fig. 8 shows the curve \(y = \frac{x}{\sqrt{x-2}}\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \((11, 11)\). \includegraphics{figure_8}

1. Verify that the \(x\)-coordinate of P is 3. [2]
2. Show that, for the curve, \(\frac{dy}{dx} = \frac{x-4}{2(x-2)^{\frac{3}{2}}}\). Hence find the gradient of the curve at P. Use the result to show that the curve is not symmetrical about \(y = x\). [7]
3. Using the substitution \(u = x - 2\), show that \(\int_3^{11} \frac{x}{\sqrt{x-2}} \, dx = 25\frac{1}{3}\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\). [9]

\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation $$y = \sin (\cos x).$$ The curve cuts the \(x\)-axis at the points \(A\) and \(C\) and the \(y\)-axis at the point \(B\).

1. Find the coordinates of the points \(A\), \(B\) and \(C\). [3]
2. Prove that \(B\) is a stationary point. [2]

Given that the region \(OCB\) is convex,

1. show that, for \(0 \leq x \leq \frac{\pi}{2}\), $$\sin (\cos x) \leq \cos x$$ and $$(1 - \frac{2}{\pi} x) \sin 1 \leq \sin (\cos x)$$ and state in each case the value or values of \(x\) for which equality is achieved. [6]
2. Hence show that $$\frac{\pi}{4} \sin 1 < \int_0^{\frac{\pi}{2}} \sin(\cos x) \, dx < 1.$$ [4]

A particle, of mass 400 grams, is initially at rest at the point \(O\). The particle starts to move in a straight line so that its velocity, \(v\) m s⁻¹, at time \(t\) seconds is given by \(v = 6t^2 - 12t^3\) for \(t > 0\)

1. Find an expression, in terms of \(t\), for the force acting on the particle. [3 marks]
2. Find the time when the particle next passes through \(O\). [5 marks]

A curve has equation $$y = x - a\sqrt{x} + b$$ where \(a\) and \(b\) are constants. The curve intersects the line \(y = 2\) at points with coordinates \((1, 2)\) and \((9, 2)\), as shown in the diagram below. \includegraphics{figure_1}

1. Show that \(a\) has the value 4 and find the value of \(b\) [3 marks]
2. On the diagram, the region enclosed between the curve and the line \(y = 2\) is shaded. Show that the area of this shaded region is \(\frac{16}{3}\) Fully justify your answer. [6 marks]

Four finite regions \(A\), \(B\), \(C\) and \(D\) are enclosed by the curve with equation $$y = x^3 - 7x^2 + 11x + 6$$ and the lines \(y = k\), \(x = 1\) and \(x = 4\), as shown in the diagram below. \includegraphics{figure_11} The areas of \(B\) and \(C\) are equal. Find the value of \(k\). [3 marks]

The diagram below shows a sketch of the curve C with equation \(y = 2 - 3x - 2x^2\) and the line L with equation \(y = x + 2\). The curve and the line intersect the coordinate axes at the points A and B. \includegraphics{figure_14}

1. Write down the coordinates of A and B. [2]
2. Calculate the area enclosed by C and L. [6]

\includegraphics{figure_17} The diagram above shows a sketch of the curve \(y = 3x - x^2\). The curve intersects the \(x\)-axis at the origin and at the point \(A\). The tangent to the curve at the point \(B(2, 2)\) intersects the \(x\)-axis at the point \(C\).

1. Find the equation of the tangent to the curve at \(B\). [4]
2. Find the area of the shaded region. [8]

The diagram below shows a sketch of the curve \(C_1\) with equation \(y = -x^2 + \pi x + 1\) and a sketch of the curve \(C_2\) with equation \(y = \cos 2x\). The curves intersect at the points where \(x = 0\) and \(x = \pi\). \includegraphics{figure_9} Calculate the area of the shaded region enclosed by \(C_1\), \(C_2\) and the \(x\)-axis. Give your answer in terms of \(\pi\). [9]

\includegraphics{figure_9} The diagram above shows a sketch of the curves \(y = x^2 + 4\) and \(y = 12 - x^2\). Find the area of the region bounded by the two curves. [6]

In this question you must show detailed reasoning. \includegraphics{figure_9} The diagram shows the curve \(y = \frac{4\cos 2x}{3 - \sin 2x}\) for \(x > 0\), and the normal to the curve at the point \((\frac{1}{4}\pi, 0)\). Show that the exact area of the shaded region enclosed by the curve, the normal to the curve and the \(y\)-axis is \(\ln \frac{2}{3} + \frac{1}{128}\pi^2\). [10]

The diagram shows the curve \(y = 4 - x^2\) and the line \(y = x + 2\). \includegraphics{figure_3}

1. Find the \(x\)-coordinates of the points of intersection of the curve and the line. [2]
2. Use integration to find the area of the shaded region bounded by the line and the curve. [6]

The diagram shows the curve \(y = 4 - x^2\) and the line \(y = x + 2\). \includegraphics{figure_6}

1. Find the \(x\)-coordinates of the points of intersection of the curve and the line. [2]
2. Use integration to find the area of the shaded region bounded by the line and the curve. [6]

\includegraphics{figure_7} Figure 7 shows the curves with equations $$y = kx^2 \quad x \geq 0$$ $$y = \sqrt{kx} \quad x \geq 0$$ where \(k\) is a positive constant. The finite region \(R\), shown shaded in Figure 7, is bounded by the two curves. Show that, for all values of \(k\), the area of \(R\) is \(\frac{1}{3}\) [5]

\includegraphics{figure_9} Figure 2 shows a sketch of part of the curve \(C\) with equation \(y = x \ln x, \quad x > 0\) The line \(l\) is the normal to \(C\) at the point \(P(e, e)\) The region \(R\), shown shaded in Figure 2, is bounded by the curve \(C\), the line \(l\) and the \(x\)-axis. Show that the exact area of \(R\) is \(Ae^2 + B\) where \(A\) and \(B\) are rational numbers to be found. [9]

The diagram shows the circle with centre O and radius 2, and the parabola \(y = \frac{1}{\sqrt{3}}(4 - x^2)\). \includegraphics{figure_5} The circle meets the parabola at points \(P\) and \(Q\), as shown in the diagram.

1. Verify that the coordinates of \(Q\) are \((1, \sqrt{3})\). [3]
2. Find the exact area of the shaded region enclosed by the arc \(PQ\) of the circle and the parabola. [8]

The curve \(y = x^3\) intersects the line \(y = kx\), \(k > 0\), at the origin and the point \(P\). The region bounded by the curve and the line, between the origin and \(P\), is denoted by \(R\).

1. Show that the area of the region \(R\) is \(\frac{1}{6}k^3\). [3]

The line \(x = a\) cuts the region \(R\) into two parts of equal area.

1. Show that \(k^3 - 6a^2k + 4a^3 = 0\). [3]

The gradient of the line \(y = kx\) increases at a constant rate with respect to time \(t\). Given that \(\frac{dk}{dt} = 2\),

1. determine the value of \(\frac{da}{dt}\) when \(a = 1\) and \(k = 2\), [4]
2. determine the value of \(\frac{da}{dt}\) when \(a = 1\) and \(k = 2\), expressing your answer in the form \(p + q\sqrt{3}\), where \(p\) and \(q\) are integers. [5]