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CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

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Question : 34 of 36
Marks: +1, -0
Using integration find the area of the region bounded between the two circles x2+y2=9x^2+y^2=9 and (x−3)2+y2=9(x-3)^2+y^2=9.
OR
Evaluate the following integral as the limit of sums ∫14(x2−x)dx\int\limits_{1}^{4} (x^2-x) dx.
Solution:     👈: Video Solution
OR
Let I=∫14(x2−x)dxI=\int\limits_{1}^{4} (x^2-x) dx
We know ∫abf(x)dx=lim⁡n→∞h[f(a)+f(a+h)+f(a+2h)+⋯+f(a+(n−1)h)]\int\limits_{a}^{b} f(x) dx = \lim\limits_{n \rightarrow \infty} h[f(a)+f(a+h)+f(a+2h)+\dots+f(a+(n-1)h)] ,
As n→∞,h→0⇒nh=b−a=4−1=3n \rightarrow \infty, h \rightarrow 0 \Rightarrow nh = b - a = 4-1=3
∴∫abf(x)dx=lim⁡n→∞h∑r=0n−1f(a+rh)      ⋯(i)\therefore \int\limits_{a}^{b} f(x) dx = \lim\limits_{n \rightarrow \infty} h \sum\limits_{r=0}^{n-1} f(a+rh) \;\;\; \cdots (i)
Here f(x)=x2−x,a=1,b=4f(x)=x^2-x, a=1, b=4 .
∴f(a+rh)=(a+rh)2−(a+rh)\therefore f(a+rh) = (a+rh)^2 - (a+rh)
⇒f(1+rh)=(1+rh)2−(1+rh)\Rightarrow f(1+rh) = (1+rh)^2 - (1+rh)
By using (i), ∫14(x2−x)dx=lim⁡n→∞h∑r=0n−1[r2h2+rh]\int\limits_{1}^{4} (x^2-x) dx = \lim\limits_{n \rightarrow \infty} h \sum\limits_{r=0}^{n-1} [r^2 h^2 + r h]
⇒I=lim⁡n→∞h{h2∑r=0n−1r2+h∑r=0n−1r}\Rightarrow I = \lim\limits_{n \rightarrow \infty} h \left\{ h^2 \sum\limits_{r=0}^{n-1} r^2 + h \sum\limits_{r=0}^{n-1} r \right\}
⇒I=lim⁡n→∞h{h2×  n(n−1)(2n−1)6+h  n(n−1)2}\Rightarrow I = \lim\limits_{n \rightarrow \infty} h \left\{ h^2 \times \; \frac{n(n-1)(2n-1)}{6} + h \; \frac{n(n-1)}{2} \right\}
⇒I=lim⁡n→∞{  nh(nh−h)(2nh−h)6+  nh(nh−h)2}\Rightarrow I = \lim\limits_{n \rightarrow \infty} \left\{ \; \frac{n h (n h - h)(2 n h - h)}{6} + \; \frac{n h (n h - h)}{2} \right\}
⇒I=  3(3−0)(6−0)6+  3(3−0)2\Rightarrow I = \; \frac{3(3-0)(6-0)}{6} + \; \frac{3(3-0)}{2}
⇒I=9+  92=  272.\Rightarrow I = 9 + \; \frac{9}{2} = \; \frac{27}{2} .
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