Question

If $h(a) = h(b),$ then $\int_{a}^{b}{\lbrack f(g\lbrack h(x)\rbrack)\rbrack^{- 1}}f'(g\lbrack h(x)\rbrack)g'\lbrack h(x)\rbrack h'(x)dx$ is equal to

• A. 0
• B. $f(a) - f(b)$
• C. $f\lbrack g(a)\rbrack - f\lbrack g(b)\rbrack$
• D. None

Answer 1

Answer: (A)

0

Answer 2

Put $f(g\lbrack h(x)\rbrack) = t$ ⇒ $f'(g\lbrack h(x)\rbrack)g'\lbrack h(x)\rbrack h'(x)dx = dt$

∴ $\int_{f(g\lbrack h(a)\rbrack)}^{f(g\lbrack h(b)\rbrack)}{t^{- 1}dt = \left\lbrack \log(t) \right\rbrack_{f(g\lbrack h(a)\rbrack)}^{f(g\lbrack h(b)\rbrack)}} = 0$ $\lbrack\because h(a) = h(b)\rbrack$