Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन1 Mark
Question
समाकलन $\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ का मान है:
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Answer
माना $I=\int_{\frac1 3}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ $=\int_{\frac1 3}^{1} \frac{\left(x^{3}\right)^{\frac1 3}\left(\frac{x}{x^{3}}-\frac{x^{3}}{x^{3}}\right)^{\frac1 3}}{x^{4}} d x$ [अंश में $\left(x^{3}\right)^{\frac1 3}$ से गुणा और भाग करने पर] $=\int_{\frac 1 3}^{1} \frac{\left(\frac{1}{x^{2}}-1\right)^{\frac1 3}}{x^{3}} d x \frac{1}{x^{2}}=t$ रखने पर, $\Rightarrow$ $\frac{-2}{x^{3}} d x=d t \Rightarrow \frac{d x}{x^{3}}=\frac{d t}{(-2)}$ सीमाओं के लिए, जब x = 1 $\Rightarrow$ t = 1 और जब $x=\frac{1}{3} \Rightarrow t=3^{2}$ = 9 ($\because$ t = $\frac{1}{x^2}$) $\therefore$ $I=\int_{9}^{1}(t-1)^{\frac1 3} \frac{d t}{(-2)}$ $=-\frac{1}{2}\left[\frac{(t-1)^{\frac{1}{3}}+1}{\frac{1}{3}+1}\right]_{9}^{1}$ $=-\frac{1}{2} \times \frac{3}{4}\left[(t-1)^{\frac{4}{3}}\right]_{9}^{1}$ $=-\frac{3}{8}\left[(1-1)^{\frac{4}{3}}-(9-1)^{\frac{4}{3}}\right]$ $=-\frac{3}{8}\left[0-\left(2^{3}\right)^{\frac{4}{3}}\right]$ $=-\frac{3}{8} \times(-16)=6$
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