Solutions for Questions 2 (page ).

Solution 2.1: Here are the answers. A * indicates that the problem involved graphs crossing the axis.

$${\textstyle\frac{31}{5}}$$,    e - e^-1,    2,    $${\textstyle\frac{7}{6}}$$,    *1,    *$${\textstyle\frac{11}{6}}$$,    *$${\textstyle\frac{17}{4}}$$,    1

Solution 2.2: Graphs meet at (- 1, 1) and at (2, 4). The area required is the area under the line minus the area under the curve for -1$\le$x$\le$2. This is $$\int_{-1}^{2}$$(x + 2 - x²) dx = 9/2.

Solution 2.3: The first graph is an upside-down parabola crossing the x-axis at 0 and 1. The second is an ordinary parabola. They meet at (0, 0) and at (1/2, 1/4). The required area is

$$\int_{0}^{1/2}$$(x(1 - x) - x²) dx = $${\textstyle\frac{1}{24}}$$

Solution 2.4: On the range 0$\le$x$\le$1 the function e^x is always $\ge$1, whilst x² is always $\le$1. So the required area is $$\int_{0}^{1}$$(e^x - x²) dx = e - $${\textstyle\frac{4}{3}}$$.

Solution 2.5: The whole area is $\int_{-1}^{1}$1 - x² dx = 4/3. The line y = 1 - a² cuts the graph at (±a, 1 - a²). The area above this line is $\int_{-a}^{a}$(1 - x²) dx - 2a(1 - a²) = ${\frac{4}{3}}$a³. So we get the required division if a³ = 1/2.

Solution 2.6:

First note that if x > 1 the $\int_{0}^{x}$b(t) dt = $\int_{0}^{1}$b(t) dt + $\int_{1}^{x}$b(t) dt = $\int_{0}^{1}$b(t) dt, i.e. the area under the graph does not change once x gets bigger than 1. The graph of b₁(x) is the straight line y = x between x = - 1 and x = 1. For x > 1 it has the constant value 1 and for x < - 1 it has the constant value -1.

Solution 2.7:The expanded integrand is g(x)²t² + 2f (x)g(x)t + f (x)². So I(t) = At² + 2Bt + C where A = $\int_{a}^{b}$g(x)² dx, B = $\int_{a}^{b}$f (x)g(x) dx and C = $\int_{a}^{b}$f (x)² dx.

Solution 2.8:The equation of the chord is y - a² = (a + b)(x - a). So the area of the segment is $\int_{a}^{b}$(a² + (a + b)(x - a) - x²) dx and this works out to give (b - a)³/6. Hence the answer.

Solution 2.9: Area under graph is 1. Putting a rectangle round the graph with opposite corners (0, 0) and ($\pi$/2, 1) and drawing a diagonal between these corners we get $\pi$/4 < 1 < $\pi$/2. So 2 < $\pi$ < 4. Similar story for the rest, with a different rectangle.

Solution 2.10: No solution.