If f(x)= ^ x _0t dt, the write the value of f'(x). - Vidyadip

Question

If $\text{f(x)}=\int\limits^{\text{x}}_0\text{t}\sin\text{t dt},$ the write the value of f'(x).

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Answer

$\text{f(x)}=\int\limits^{\text{x}}_0\text{t}\sin\text{t dt}$
$\Rightarrow\text{f(x)}=\text{t}\big[-\cos\text{t}\big]^{\text{x}}_0-\int\limits^{\text{x}}_0\frac{\text{d}}{\text{dt}}(\text{t})\times(-\cos\text{t})\text{dt}$
$\Rightarrow\text{f(x)}=-(\text{x}\cos\text{x}-0)+\int\limits^{\text{x}}_0\cos\text{t dt}$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+\big[\sin\text{t}\big]^{\text{x}}_0$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+(\sin\text{x}-0)$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+\sin\text{x}$
Differentiating both sides with respect to x, we get
$\text{f}'(\text{x})=-\big[\text{x}\times(-\sin\text{x})+\cos\text{x}\times1\big]+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=-(-\text{x}\sin\text{x})-\cos\text{x}+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\text{x}\sin\text{x}$
Thus, the value of f'(x) is x $\sin\text{x}$

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