Question

Let f be a continuous function on the interval [a,b], and let c = a+b 2 . Suppose that F is any antiderivative of f with F(a) = 3 and F(b) = 7.
Find: (a)Rb a f(x)dx (2 marks) (b) −Ra b f(x)dx (2 marks) (c)Rc a f(x)dx +Rb c f(x)dx (2 marks) (d) Dx[−(Rc a f(x)dx +Rb c f(x)dx)] (2 marks) 3. Find the area bounded by the curve y = |x−2|, the x-axis and the lines x = −7 and x = 11. (5 marks)

Answer 1

Let $f$ be a continuous function on the interval $[a,b],$ and let $c=\dfrac{a+b}{2}.$ Suppose that $F$ is any antiderivative of $f$ with $F(a)=3$ and $F(b)=7.$

Find:

$$\displaystyle\int_{a}^bf(x)dx$$

By the Fundamental Theorem of Calculus (Part 2) if $f$  is continuous on $[a, b]$  then

$$\displaystyle\int_{a}^bf(x)dx=F(b)-F(a),$$

where $F$ is any antiderivative of $f.$

Given $F(a)=3$ and $F(b)=7.$ Then

$$\displaystyle\int_{a}^bf(x)dx=F(b)-F(a)=7-3=4$$

b)

$$-\displaystyle\int_{b}^af(x)dx=-(-\displaystyle\int_{a}^bf(x)dx)=\displaystyle\int_{a}^bf(x)dx=4$$

c)

$$\displaystyle\int_{a}^cf(x)dx+\displaystyle\int_{c}^bf(x)dx=\displaystyle\int_{a}^bf(x)dx=4$$

d)

$$D_x[-(\displaystyle\int_{a}^cf(x)dx+\displaystyle\int_{c}^bf(x)dx)]=D_x[-4]=0$$

2.

$$y=|x-2|= \begin{cases} x-2 &\text{if } x\geq 2 \\ -x+2 &\text{if } x<2 \end{cases}$$

$$Area=A=\displaystyle\int_{-7}^2(-x+2)dx+\displaystyle\int_{2}^{11}(x-2)f(x)dx=$$

$$=[-{x^2\over 2}+2x]\begin{matrix} 2 \\ -7 \end{matrix}+[{x^2\over 2}-2x]\begin{matrix} 11 \\ 2 \end{matrix}=$$

$$=-2+4+{49\over 2}+14+{121\over 2}-22-2+4=81(units^2)$$

Area is $81$ square units.