12th NCERT INTEGRALS Exercise 7.8 Questions 6
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Evaluate the following definite integrals as limit of sums

Question (1)

\[\int_a^b {x\;dx} \]

Solution

Divide (a, b) into n equal sub intervals of length "h" each then h=(b-a)/n
Here a =a , b= b
f(x) = x
f(a+ih) = (a+ih)
By definition \[I = \mathop {\lim }\limits_{h \to \infty } \;h \cdot \sum\limits_{i = 1}^n {f\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to \infty } \;h \cdot \sum\limits_{i = 1}^n {\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to \infty } \;ah\;\sum\limits_{i = 1}^n 1 + \mathop {\lim }\limits_{h \to \infty } \;{h^2}\;\sum\limits_{i = 1}^n i \] \[I = \mathop {\lim }\limits_{h \to \infty } \;ah \cdot n + \mathop {\lim }\limits_{h \to \infty } \;{h^2} \cdot \frac{{n\left( {n + 1} \right)}}{2}\] \[ = \mathop {\lim }\limits_{h \to \infty } a\frac{{\left( {b - a} \right)}}{\require{cancel} \cancel{n}} \cdot \cancel{n} + \mathop {\lim }\limits_{h \to \infty } {\left( {\frac{{b - a}}{n}} \right)^2} \cdot \frac{{n\left( {n + 1} \right)}}{2}\] as h → 0 n → ∞ \[I = ab - {a^2} + {\left( {b - a} \right)^2}\mathop {\lim }\limits_{n \to \infty } \;\frac{1}{{{n^\cancel{2}}}}\frac{{\cancel{n}\left( {n + 1} \right)}}{2}\] \[I = ab - {a^2} + \frac{{\left( {{b^2} - 2ab + {a^2}} \right)}}{2}\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right)\] as n→ ∞ (1/n) → 0 \[I = ab - {a^2} + \frac{{\left( {{b^2} - 2ab + {a^2}} \right)}}{2}\left( 1 \right)\] \[I = \frac{{\cancel{2ab} - 2{a^2} + {b^2} \cancel{- 2ab} + {a^2}}}{2}\] \[I = \frac{{{b^2} - {a^2}}}{2}\]

Question (2)

\[\int_0^5 {\left( {x + 1} \right)dx} \]

Solution

Here a = 0, b=5
f(x) = x+1
f(a+ih)= f(0+ih)
f(a+ih)= f(ih)= ih+1
divide nterval (0 - 5) in "n" equal subintervals of length "h" each \[h = \frac{{b - a}}{n} = \frac{{5 - 0}}{n} = \frac{5}{n}\]
By definition \[I = \mathop {\lim }\limits_{h \to \infty } h\sum\limits_{i = 1}^n {f\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to \infty } h\sum\limits_{i = 1}^n {\left( {ih + 1} \right)} \] \[I = \mathop {\lim }\limits_{h \to \infty } {h^2}\sum\limits_{i = 1}^n i + \mathop {\lim }\limits_{h \to \infty } h\sum\limits_i^n 1 \] as h → 0 , n → ∞
\[I = \mathop {\lim }\limits_{n \to \infty } \frac{{25}}{{{n^\cancel{2}}}} \cdot \frac{{\cancel{n}\left( {n + 1} \right)}}{2} + \mathop {\lim }\limits_{n \to \infty } \frac{5}{\cancel{n}} \cdot \cancel{n}\] \[I = \frac{{25}}{2}\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right) + 5\] \[I = \frac{{25}}{2} + 5\] \[I = \frac{{35}}{2}\]

Question (3)

\[I = \int_2^3 {{x^2}} dx\]

Solution

Here a =2 b =3
f(x) = x2
f(a+ih) = f(2+ih)
f(a+ih) = (2+ih)2
f(a+ih)=4+4ih+i2h2
Divide (2, 3) in n equal sub intervals of leength 'h' each
\[h = \frac{{3 - 2}}{n} = \frac{1}{n}\] By definition \[I = \mathop {\lim }\limits_{h \to o} \;h\sum\limits_{i = 1}^n {f\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to o} \;h\sum\limits_{i = 1}^n {\left( {4 + 4ih + {i^2}{h^2}} \right)} \] \[I = \mathop {\lim }\limits_{h \to 0} 4h\sum\limits_{i = 1}^n 1 + \mathop {\lim }\limits_{h \to 0} 4{h^2}\sum\limits_{i = 1}^n i + \mathop {\lim }\limits_{h \to 0} {h^3}\sum\limits_{i = 1}^n {{i^2}} \] as h → 0 n → ∞ \[I = \mathop {\lim }\limits_{n \to \infty } 4hn + \mathop {\lim }\limits_{n \to \infty } 4{h^2}\frac{{n\left( {n + 1} \right)}}{2} + \mathop {\lim }\limits_{n \to \infty } {h^3}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] \[I = \mathop {\lim }\limits_{n \to \infty } 4\frac{1}{\cancel{n}}\cancel{n} + \mathop {\lim }\limits_{n \to \infty } 4\frac{1}{{{n^\cancel {2}}}}\frac{{\cancel{n}\left( {n + 1} \right)}}{2} + \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3}}}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] \[I = 4 + \frac{\cancel{4}}{\cancel{2}}\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right) + \frac{1}{6}\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right)\left( {2 + \frac{1}{n}} \right)\] \[I = 4 + 2\left( 1 \right) + \frac{1}{\cancel{6}3}\left( 1 \right)\left( \cancel{2} \right)\] \[I = 6 + \frac{1}{3} = \frac{{19}}{3}\]

Question (4)

\[I = \int_1^4 {\left( {{x^2} - x} \right)dx} \]

Solution

a = 1 n b = 4
f(x) = x2 - x
f(a+ih) =f(1+ih)
f(a+ih) = (1+ih)2 - (1+ih)
f(a+ih) = 1 +2ih +i2h2 - 1 - ih
f(a+ih) = +2ih +i2h2 - ih
f(a+ih) = i2h2 + ih
Divide (1, 4) in n equal sub intervals of length h each \[h = \frac{{4 - 1}}{n} = \frac{3}{n}\] as h → 0, n → ∞
By definition I \[I = \mathop {\lim }\limits_{h \to 0} \,h\sum {f\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to 0} \,h\sum {\left( {{i^2}{h^2} + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to 0} \,{h^3}\sum {{i^2} + } \mathop {\lim }\limits_{h \to 0} \,{h^2}\sum i \] as h → , n → ∞ Replacing value of h and ∑i2, ∑i we get \[I = \mathop {\lim }\limits_{n \to \infty } \,{\left( {\frac{3}{n}} \right)^3} \cdot \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \mathop {\lim }\limits_{n \to \infty } \frac{9}{{{n^\cancel{2}}}}\frac{{\cancel{n}\left( {n + 1} \right)}}{2}\] \[I = \frac{{\cancel{27}^9}}{\cancel{6}_2}\mathop {\lim }\limits_{n \to \infty } \,\frac{{\cancel{n} \times \cancel{n}\left( {1 + \frac{1}{n}} \right) \times \cancel{n}\left( {2 + \frac{1}{n}} \right)}}{{{\cancel{n^3}}}} + \frac{9}{2}\mathop {\lim }\limits_{n \to \infty } \frac{{\cancel{n} \times \cancel{n}\left( {1 + \frac{1}{n}} \right)}}{{{\cancel{n^2}}}}\] \[I = \frac{9}{\cancel{2}}\left(\cancel{2} \right) + \frac{9}{2}\left( 1 \right)\] \[I = 9 + \frac{9}{2}\] \[I = \frac{{27}}{2}\]

Question (5)

\[I = \int_{ - 1}^1 {{e^x}} dx\]

Solution

Let a = -1, b = 1
Divide (-1, 1) in nequal parts of length 'h' each \[h = \frac{{1 - \left( { - 1} \right)}}{n} = \frac{2}{n}\] f(x) = ex
f(a+ih) = f(-1+ih)
f(a+ih) = e-1+ih f(a+ih) = e-1+eih By definition \[I = \mathop {\lim }\limits_{h \to o} h\;\sum\limits_{i = 1}^n {f\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to o} h\;\sum\limits_{i = 1}^n {{e^{ - 1}} \cdot {e^{ih}}} \] \[I = \mathop {\lim }\limits_{h \to o} \frac{h}{e}\;\sum\limits_{i = 1}^n {{e^{ih}}} \] \[I = \mathop {\lim }\limits_{h \to o} \frac{h}{e}\;\left[ {{e^h} + {e^{2h}} + {e^{3h}} + .... + {e^{nh}}} \right]\] It is G.P. with a = eh, r = eh > 1 So \[{S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\] as h→ 0 \[I = \mathop {\lim }\limits_{h \to 0} \frac{h}{e} \cdot \frac{{{e^h}\left[ {{{\left( {{e^h}} \right)}^n} - 1} \right]}}{{{e^n} - 1}}\] \[I = \frac{1}{e}\mathop {\lim }\limits_{h \to 0} h \cdot \frac{{{e^h}\left[ {{{\left( {{e^h}} \right)}^n} - 1} \right]}}{{{e^n} - 1}}\] Divide by h to denominator and numerator \[I = \frac{1}{e}\mathop {\lim }\limits_{h \to 0} \frac{{{e^h}\left[ {{{\left( {{e^h}} \right)}^n} - 1} \right]}}{{\frac{{{e^n} - 1}}{h}}}\] \[I = \frac{1}{e} \cdot \frac{{{e^0}\left( {{e^2} - 1} \right)}}{1}\] \[I = \frac{{\left( {{e^2} - 1} \right)}}{e}\] \[I = e - \frac{1}{e}\]

Question (6)

\[I = \int_0^4 {\left( {x + {e^{2x}}} \right)dx} \]

Solution

a = 0, b = 4
f(x) = x + e2x
f(a+ih) = f(0 +ih) = f(ih) f(a+ih) = ih + e2ih Divide (0, 4) in 'n' equal sub interval of length 'h' each
\[\therefore h = \frac{{4 - 0}}{n} = \frac{4}{n}\] By definition \[I = \mathop {\lim }\limits_{h \to 0 } h\;\sum\limits_{i = 1}^n {f\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to 0 } h\;\sum\limits_{i = 1}^n {\left( {ih + {e^{2ih}}} \right)} \] \[I = \mathop {\lim }\limits_{h \to 0} {h^2}\sum\limits_{i = 1}^n i + \mathop {\lim }\limits_{h \to 0} h\sum\limits_{i = 1}^n {{e^{2ih}}} \] as h → 0, n → ∞ \[I = \mathop {\lim }\limits_{n \to \infty } \frac{{\cancel{16}^8}}{{{n^\cancel{2}}}} \cdot \frac{{\cancel{n}\left( {n + 1} \right)}}{\cancel{2}} + \mathop {\lim }\limits_{h \to 0} h\left[ {{e^{2h}} + {e^{4h}} + {e^{6h}} + ... + {e^{2nh}}} \right]\] \[I = 8\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right) + \mathop {\lim }\limits_{h \to 0} h\left[ {{e^{2h}} + {e^{4h}} + {e^{6h}} + ... + {e^{2nh}}} \right]\] e2h + e4h +...+ e2nh is G.P. a= e2h, r = e2h \[I = 8 + \mathop {\lim }\limits_{h \to 0} h\frac{{{e^{2h}}\left[ {{{\left( {{e^{2h}}} \right)}^n} - 1} \right]}}{{{e^{2h}} - 1}}\] Divide denominator and numerator by 'h' \[I = 8 + \mathop {\lim }\limits_{h \to 0} h\frac{{{e^{2h}}\left[ {{e^{2nh}} - 1} \right]}}{{2 \times \frac{{{e^{2h}} - 1}}{{2h}}}}\] Substitute value of nh = 4 already proved \[I = 8 + \frac{{{e^0}\left( {{e^8} - 1} \right)}}{{2 \times 1}}\] \[I = 8 + \frac{{{e^8} - 1}}{2}\] \[I = \frac{{{e^8} + 15}}{2}\]
Exercise 7.7⇐
⇒ Exercise 7.9