Assertion (A): int_{-a}^a f(x) dx = int_0^a ( | vedclass

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(C) For Assertion $(A)$: We know that $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$. In the first integral,let $x = -t$,then $dx = -d

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$(A)$ is true but $(R)$ is false

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(C) For Assertion $(A)$: We know that $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$. In the first integral,let $x = -t$,then $dx = -dt$. When $x = -a, t = a$ and when $x = 0, t = 0$. So,$\int_{-a}^0 f(x) dx = \int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt = \int_0^a f(-x) dx$. Thus,$\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx = \int_0^a (f(x) + f(-x)) dx$. Hence,$(A)$ is true. For Reason $(R)$: The substitution rule for definite integrals states that if $x = g(u)$,then $dx = g'(u) du$. The limits change from $a$ to $b$ to $g^{-1}(a)$ to $g^{-1}(b)$. The formula provided in $(R)$ is $\int_a^b f(x) dx = \int_{g(a)}^{g(b)} f(g(u)) g'(u) du$,which is incorrect because the limits of integration on the right side should be $g^{-1}(a)$ and $g^{-1}(b)$,not $g(a)$ and $g(b)$. Thus,$(R)$ is false.

Assertion $(A): \int_{-a}^a f(x) dx = \int_0^a (f(x) + f(-x)) dx$
Reason $(R): \int_a^b f(x) dx = \int_{g(a)}^{g(b)} f(g(u)) g'(u) du$
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Assertion $(A): \int_{-a}^a f(x) dx = \int_0^a (f(x) + f(-x)) dx$Reason $(R): \int_a^b f(x) dx = \int_{g(a)}^{g(b)} f(g(u)) g'(u) du$The correct option among the following is: