Area between curve and line

\includegraphics{figure_2} Figure 2 shows part of the curve \(C\) with equation $$y = 9 - 2x - \frac{2}{\sqrt{x}}, \quad x > 0.$$ The point \(A(1, 5)\) lies on \(C\) and the curve crosses the \(x\)-axis at \(B(b, 0)\), where \(b\) is a constant and \(b > 0\).

1. Verify that \(b = 4\). [1]

The tangent to \(C\) at the point \(A\) cuts the \(x\)-axis at the point \(D\), as shown in Fig. 2.

1. Show that an equation of the tangent to \(C\) at \(A\) is \(y + x = 6\). [4]
2. Find the coordinates of the point \(D\). [1]

The shaded region \(R\), shown in Fig. 2, is bounded by \(C\), the line \(AD\) and the \(x\)-axis.

1. Use integration to find the area of \(R\). [6]

\includegraphics{figure_1} Figure 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_7} 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_2} Figure 2 shows the curve \(C\) with equation \(y = 3x - 4\sqrt{x} + 2\) and the tangent to \(C\) at the point \(A\). Given that \(A\) has \(x\)-coordinate 4,

1. show that the tangent to \(C\) at \(A\) has the equation \(y = 2x - 2\). [6]

The shaded region is bounded by \(C\), the tangent to \(C\) at \(A\) and the positive coordinate axes.

1. Find the area of the shaded region. [8]

\includegraphics{figure_9} The diagram shows the curve \(y = 2x^2 + 6x + 7\) and the straight line \(y = 2x + 13\).

1. Find the coordinates of the points where the curve and line intersect. [4]
2. Show that the area of the shaded region bounded by the curve and line is given by $$\int_{-3}^{1} (6 - 4x - 2x^2) dx.$$ [2]
3. Hence find the area of the shaded region. [5]

\includegraphics{figure_7} The diagram shows the curve with equation \(y = (3x - 1)^4\). The point P on the curve has coordinates \((1, 16)\) and the tangent to the curve at P meets the \(x\)-axis at the point Q. The shaded region is bounded by PQ, the \(x\)-axis and that part of the curve for which \(\frac{1}{3} \leqslant x \leqslant 1\). Find the exact area of this shaded region. [10]

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]

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]

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 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]

In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. \includegraphics{figure_2} Figure 2 Figure 2 shows a sketch of part of the curve \(C\) with equation $$y = x^3 - 10x^2 + 27x - 23$$ The point \(P(5, -13)\) lies on \(C\) The line \(l\) is the tangent to \(C\) at \(P\)

1. Use differentiation to find the equation of \(l\), giving your answer in the form \(y = mx + c\) where \(m\) and \(c\) are integers to be found. [4]
2. Hence verify that \(l\) meets \(C\) again on the \(y\)-axis. [1]

The finite region \(R\), shown shaded in Figure 2, is bounded by the curve \(C\) and the line \(l\).

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

\includegraphics{figure_1} The diagram shows the curve \(y = 6x - x^2\) and the line \(y = 5\). Find the area of the shaded region. You must show detailed reasoning. [4]

\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]