CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

Question : 34 of 36

Marks: +1, -0

Using integration find the area of the region bounded between the two circles

x^2+y^2=9

and

(x-3)^2+y^2=9

.
OR
Evaluate the following integral as the limit of sums

∫↙{1}↖{4} (x^2-x) {\d} x

Solution:

OR

Let

I=∫↙{1}↖{4} (x^2-x) \dx

We know

∫↙{a}↖{b} f(x) \dx=\lim ↙{n → ∞} h[f(a)+f(a+h)+f(a+2 h)+. . .+f(a+(n-1) h)]

,
As

{\n} → ∞, {\h} → 0 ⇒ {\nh}= {\b}- {\a}=4-1=3

∴ ∫↙{ {\a}}↖{ {\b}} {\f}( {\x}) {\dx}=\lim ↙{ {\n} → ∞} {\h} ∑↙{{ {\r}=0}}↖{{ {\n}}-1} {\f}( {\a}+ {\rh}) \;\;\;⋅⋅⋅⋅⋅⋅⋅(i)

Here

f(x)=x^2-x, a=1, b=4

∴ f(a+r h)=(a+r h)^2-(a+r h)

⇒ f(1+r h)=(1+r h)^2-(1+r h)

By using (i),

∫↙{1}↖{4} (x^2-x) \dx=\lim ↙{n → ∞} h ∑↙{{r=0}}↖{{n-1}} [r^2 h^2+r h]

⇒ I=\lim ↙{n → ∞} h \{h^2 ∑↙{{r=0}}↖{{n-1}} r^2+h ∑↙{{r=0}}↖{{n-1}} r\}

⇒ I=\lim ↙{n → ∞} h \{h^2 × \;{n(n-1)(2 n-1)}/{6}+h \;{n(n-1)}/{2}\}

⇒ I=\lim ↙{n → ∞} \{\;{n h(n h-h)(2 n h-h)}/{6}+\;{n h(n h-h)}/{2}\}

⇒ I=\;{3(3-0)(6-0)}/{6}+\;{3(3-0)}/{2}

⇒ I=9+\;{9}/{2}=\;{27}/{2} .

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