If h(a) = h(b),the value of the integral int_ | vedclass

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(A) Let $u = h(x)$. Then $du = h'(x) \, dx$.When $x = a$,$u = h(a)$. When $x = b$,$u = h(b)$.The integral becomes $\int_{h(a)}^{h(b)} [f(g(u))]^{-1} f

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $u = h(x)$. Then $du = h'(x) \, dx$.When $x = a$,$u = h(a)$. When $x = b$,$u = h(b)$.The integral becomes $\int_{h(a)}^{h(b)} [f(g(u))]^{-1} f'(g(u)) g'(u) \, du$.Let $v = g(u)$. Then $dv = g'(u) \, du$.When $u = h(a)$,$v = g(h(a))$. When $u = h(b)$,$v = g(h(b))$.The integral becomes $\int_{g(h(a))}^{g(h(b))} [f(v)]^{-1} f'(v) \, dv$.This is $\int_{g(h(a))}^{g(h(b))} \frac{f'(v)}{f(v)} \, dv = [\ln|f(v)|]_{g(h(a))}^{g(h(b))}$.Since $h(a) = h(b)$,the upper and lower limits are equal: $g(h(a)) = g(h(b))$.Therefore,the value of the integral is $0$.

If $h(a) = h(b)$,the value of the integral $\int_a^b {[f(g(h(x)))]^{-1} f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \, dx} = $

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