Which of the following inequalities is/are TR | vedclass

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(C) Using the Taylor series expansion,$\cos x \geq 1 - \frac{x^2}{2}$ for $x \in [0, 1]$.$\int_0^1 x \cos x \, dx \geq \int_0^1 x(1 - \frac{x^2}{2}) \

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$A, B, D$

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(C) Using the Taylor series expansion,$\cos x \geq 1 - \frac{x^2}{2}$ for $x \in [0, 1]$.$\int_0^1 x \cos x \, dx \geq \int_0^1 x(1 - \frac{x^2}{2}) \, dx = [\frac{x^2}{2} - \frac{x^4}{8}]_0^1 = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}$. Thus,$(A)$ is $TRUE$.$(B)$ Using $\sin x \geq x - \frac{x^3}{6}$,we have:$\int_0^1 x \sin x \, dx \geq \int_0^1 x(x - \frac{x^3}{6}) \, dx = [\frac{x^3}{3} - \frac{x^5}{30}]_0^1 = \frac{1}{3} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10}$. Thus,$(B)$ is $TRUE$.$(C)$ Since $\cos x $(D)$ Using $\sin x \geq x - \frac{x^3}{6}$,we have:$\int_0^1 x^2 \sin x \, dx \geq \int_0^1 x^2(x - \frac{x^3}{6}) \, dx = [\frac{x^4}{4} - \frac{x^6}{36}]_0^1 = \frac{1}{4} - \frac{1}{36} = \frac{9-1}{36} = \frac{8}{36} = \frac{2}{9}$. Thus,$(D)$ is $TRUE$.Therefore,the correct options are $(A), (B), (D)$.

Which of the following inequalities is/are $TRUE$?
$(A)$ $\int_0^1 x \cos x \, dx \geq \frac{3}{8}$
$(B)$ $\int_0^1 x \sin x \, dx \geq \frac{3}{10}$
$(C)$ $\int_0^1 x^2 \cos x \, dx \geq \frac{1}{2}$
$(D)$ $\int_0^1 x^2 \sin x \, dx \geq \frac{2}{9}$

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Which of the following inequalities is/are $TRUE$?$(A)$ $\int_0^1 x \cos x \, dx \geq \frac{3}{8}$$(B)$ $\int_0^1 x \sin x \, dx \geq \frac{3}{10}$$(C)$ $\int_0^1 x^2 \cos x \, dx \geq \frac{1}{2}$$(D)$ $\int_0^1 x^2 \sin x \, dx \geq \frac{2}{9}$