Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
If $f(x)=\int_{0}^{x} t \sin t d t$, then f'(x) is
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Answer
Given: $f(x)=\int_{0}^{x} t \sin t d t$ Applying product rule, $\Rightarrow \int \mathrm{u} . \mathrm{v} \mathrm{d} \mathrm{x}=\mathrm{u} \cdot \int \mathrm{vdx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$ So, $f(x)=[t]_{0}^{x} \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \cdot \int \sin t d t\right\} d t$ $=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cos t) d t$ $=[-t(\cos t)+\sin t]_{0}^{x}$ = -x cos x + sin x - 0 ⇒ f(x) = -x cos x + sinx $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\left[\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}+\cos \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}\right]$ ⇒ f'(x) = -[{x(-sinx )} + cosx ] + cosx = x sin x - cos x + cos x = x sin x
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