Question
Download Solution PDFपरवलय y = 3x2 और x2 - y + 4 = 0 से घिरा क्षेत्रफल है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
परवलय y = 3x2 और x2 - y + 4 = 0
संकल्पना:
दो वक्रों y1 और y2 के बीच के क्षेत्रफल की संकल्पना को x = a और x = b के बीच लागू करने पर
\(\rm A=\int_a^b(y_1-y_2)\ dx\)
गणना:
परवलय y = 3x2 और x2 - y + 4 = 0
तब 3x2 = x2 + 4
⇒ x2 = 2
⇒ x = ± √ 2
तब क्षेत्रफल है
\(\rm A=\int_{-\sqrt2}^{\sqrt2}(x^2+4-3x^2) \ dx\)
\(\rm A=\int_{-\sqrt2}^{\sqrt2}(4-2x^2) \ dx\)
\(\rm A=[4x-2\frac{x^3}{3}]_{-\sqrt2}^{\sqrt2}\)
\(\rm A=4[\sqrt2-(-\sqrt2)]-\frac{2}{3}[{\sqrt2}^3-{(-\sqrt2)}^3]\)
\(\rm A=8\sqrt2-\frac{8}{3}\sqrt2\)
\(\rm A=\frac{16\sqrt2}{3}\) वर्ग इकाई
अतः विकल्प (4) सही है।
Last updated on Jun 19, 2025
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