Application of Integrals — Flashcards
Madhya Pradesh Board · Class 12 · Mathematics
60 flashcards for Application of Integrals (Madhya Pradesh Board Class 12 Mathematics) to test yourself on key terms and facts.
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Find the area under the curve y = f(x) from x = a to x = b using vertical strips.
Answer
Use: A = ∫ab dA = ∫ab y dx = ∫ab f(x) dx. Steps: 1. Take a thin vertical strip of width dx. 2. Its height is y = f(x). 3. So elementary area dA = y dx. 4. Add all strips from x = a to x = b. Answer: A…
ऊर्ध्वाधर पट्टियों का उपयोग करके x = a से x = b तक वक्र y = f(x) के नीचे का क्षेत्रफल ज्ञात कीजिए।
Answer
उपयोग करें: A = ∫ab dA = ∫ab y dx = ∫ab f(x) dx. चरण: 1. चौड़ाई dx वाली एक पतली ऊर्ध्वाधर पट्टी लें। 2. इसकी ऊँचाई y = f(x) है। 3. अतः सूक्ष्म क्षेत्रफल dA = y dx. 4. x = a से x = b तक सभी पट्टियों को…
क्षैतिज पट्टियों का उपयोग करके y = c से y = d तक वक्र x = g(y) के नीचे का क्षेत्रफल ज्ञात कीजिए।
Answer
उपयोग करें: A = ∫cd x dy = ∫cd g(y) dy. चरण: 1. मोटाई dy वाली एक पतली क्षैतिज पट्टी लें। 2. इसकी लंबाई x = g(y) है। 3. अतः पट्टी का क्षेत्रफल x dy है। 4. y = c से y = d तक सभी पट्टियों को जोड़ें। उत्त…
Find the area under the curve x = g(y) from y = c to y = d using horizontal strips.
Answer
Use: A = ∫cd x dy = ∫cd g(y) dy. Steps: 1. Take a thin horizontal strip of thickness dy. 2. Its length is x = g(y). 3. So strip area is x dy. 4. Add all strips from y = c to y = d. Answer: A = ∫cd g(y…
Why is the elementary area of a thin vertical strip written as dA = y dx?
Answer
A thin vertical strip has: 1. Height = y 2. Width = dx 3. Area of a rectangle = height × width So, dA = y dx. This is the small area piece that is later added by integration.
एक पतली ऊर्ध्वाधर पट्टी का सूक्ष्म क्षेत्रफल dA = y dx के रूप में क्यों लिखा जाता है?
Answer
एक पतली ऊर्ध्वाधर पट्टी के लिए: 1. ऊँचाई = y 2. चौड़ाई = dx 3. आयत का क्षेत्रफल = ऊँचाई × चौड़ाई अतः, dA = y dx. यह क्षेत्रफल का वह छोटा भाग है जिसे बाद में समाकलन द्वारा जोड़ा जाता है.
वृत्त x² + y² = a² से घिरा क्षेत्रफल ज्ञात कीजिए।
Answer
चरण 1: वृत्त दोनों अक्षों के बारे में सममित है। चरण 2: पहले चतुर्थांश में क्षेत्रफल ज्ञात कीजिए और 4 से गुणा कीजिए। चरण 3: पहले चतुर्थांश में, y = √(a² - x²) है। चरण 4: क्षेत्रफल = 4 ∫0a √(a² - x²) dx…
Find the area enclosed by the circle x² + y² = a².
Answer
Step 1: The circle is symmetric about both axes. Step 2: Find area in first quadrant and multiply by 4. Step 3: In first quadrant, y = √(a² - x²). Step 4: Area = 4 ∫0a √(a² - x²) dx. Step 5: Use the s…
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