Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन2 Marks
Question
$\int_{-1}^{2}\left|x^{3}-x\right| d x$ का मान ज्ञात कीजिए।
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Answer
हम देखते हैं कि $[-1, 0]$ पर $x^{3 }- x \geq 0$ और $[0, 1]$ पर $x^{3 }- x \leq 0$ और$ [1, 2] $ पर $x^{3 }- x \geq 0$ तब हम लिख सकते हैं कि
$\int_{-1}^{2}\left|x^{3}-x\right| d x =\int_{-1}^{0}\left(x^{3}-x\right) d x +\int_{0}^{1}-\left(x^{3}-x\right) d x+\int_{1}^{2}\left(x^{3}-x\right)dx$
$=\int_{-1}^{0}\left(x^{3}-x\right) d x+\int_{0}^{1}\left(x-x^{3}\right) d x+\int_{1}^{2}\left(x^{3}-x\right)dx$
$=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0} +\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1} +\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}$
$=-\left(\frac{1}{4}-\frac{1}{2}\right) +\left(\frac{1}{2}-\frac{1}{4}\right) + (4 − 2) −\left(\frac{1}{4}-\frac{1}{2}\right)$
$=-\frac{1}{4}+\frac{1}{2} +\frac{1}{2}-\frac{1}{4} +2-\frac{1}{4} +\frac{1}{2}= \frac{3}{2}- \frac{3}{4}+2 =\frac{11}{4}$
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