Answer to Question #120223 in Calculus for Yola

Question

2. 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) = 1 and F(b) = −2. Find: (a) ba f(x)dx (2 marks) (b) −abf(x)dx (2 marks) (c) ca f(x)dx + bcf(x)dx (2 marks) (d) Dx[−(caf(x)dx + bcf(x)dx)] (2 marks)

Answer 1

Since F is antiderivatve of f ,

∫ f(x) dx = F(x) + K , K is integration constant.

Also as c = (a+b)/2 , a < c < b

Given here F(a) = 1 and F(b) = -2

Now ,

$\int_{a}^{b}f(x)dx$

= [F(x)]_a^b

= F(b) - F(a)

= -2-1

= -3

a)

$\int_{a}^{b}f(x)dx = F(b) - F(a) = -3$

b)

$- \int_{b}^{a }f(x)dx = \int_{a}^{b}f(x)dx$

= −3

c) $\int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx$

= $\int_{a}^{b}f(x)dx$ as a < c < b

= −3

d)

Dx[ - ( $\int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx$ )]

= Dx[-$\int_{a}^{b}f(x)dx$ ]

= Dx[-(-3)]

= D(3x)

= 3