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Question:
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Assertion
$\int_{-a}^a f(x) d x=\int_0^a(f(x)+f(-x)) d x$
Reason (R) $\int_a^b f(x) d x=\int_{g(a)}^{g(b)} f(g(u)) g^{\prime}(u) d u$
The correct option among the following is
Options:
$\int_{-a}^a f(x) d x=\int_0^a(f(x)+f(-x)) d x$
Reason (R) $\int_a^b f(x) d x=\int_{g(a)}^{g(b)} f(g(u)) g^{\prime}(u) d u$
The correct option among the following is
Solution:
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Verified Answer
The correct answer is:
$(A)$ is true but $(R)$ is false
We have, $A=\int_{-a}^a f(x) d x=\int_0^a(f(x)+f(-x)) d x$
A is true, $\quad R=\int_a^b f(x) d x=\int_{g(a)}^{g(b)} f\left(g(u) g^{\prime}(u) d u\right.$
Put, $x=g(u), d x=g(u) d u$
at $x=a=g(u)$ and $x=b \quad u=g^{-1}(b) \Rightarrow u=g^{-1}(a)$
$\therefore \quad \int_a^b f(x) d x=\int_{g^{-1}(a)}^{g^{-1}(b)} f\left(g(x) g^{\prime}(x) d x\right.$
$\therefore \mathrm{R}$ is false.
A is true, $\quad R=\int_a^b f(x) d x=\int_{g(a)}^{g(b)} f\left(g(u) g^{\prime}(u) d u\right.$
Put, $x=g(u), d x=g(u) d u$
at $x=a=g(u)$ and $x=b \quad u=g^{-1}(b) \Rightarrow u=g^{-1}(a)$
$\therefore \quad \int_a^b f(x) d x=\int_{g^{-1}(a)}^{g^{-1}(b)} f\left(g(x) g^{\prime}(x) d x\right.$
$\therefore \mathrm{R}$ is false.
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