Match the following: List-IList-III. int_{-1} | vedclass

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(I-D, II-A, III-E, IV-C) The correct matches are as follows: $I. \int_{-1}^1 x|x| dx = 0$ (since $f(x) = x|x|$ is an odd function,i.e.,$f(-x) = -x|-x|

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(I-D, II-A, III-E, IV-C) The correct matches are as follows: $I. \int_{-1}^1 x|x| dx = 0$ (since $f(x) = x|x|$ is an odd function,i.e.,$f(-x) = -x|-x| = -x|x| = -f(x)$). Thus,$I ightarrow (d)$. $II. 	ext{Let } I = \int_0^{\pi/2} \left(1 + \log \frac{4+3\sin x}{4+3\cos x}ight) dx$. Using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi/2} \left(1 + \log \frac{4+3\cos x}{4+3\sin x}ight) dx$. Adding the two expressions for $I$: $2I = \int_0^{\pi/2} \left(2 + \log \left(\frac{4+3\sin x}{4+3\cos x} \cdot \frac{4+3\cos x}{4+3\sin x}ight)ight) dx = \int_0^{\pi/2} (2 + \log 1) dx = \int_0^{\pi/2} 2 dx = \pi$. Therefore,$I = \frac{\pi}{2}$. Thus,$II ightarrow (a)$. $III. \int_0^a f(x) dx = \int_0^a f(a-x) dx$ is a standard property of definite integrals. Thus,$III ightarrow (e)$. $IV. \int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$. Let $x = -t$ in the first integral,$dx = -dt$. $int_{-a}^0 f(x) dx = \int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt = \int_0^a f(-x) dx$. So,$\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$. Thus,$IV ightarrow (c)$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

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