Integrals NCERT Solutions
NCERT Exercise 7.8
Question: 1. `int_a^b xdx`
Solution:
Here, `f(x)=x, nh =b-a`
We know that, `int_a^b f(x)dx`
`=lim_(h->0) h[f(a)+f(a+h)+` `f(a+2h) ....+f(a+bar(n-1)h)]`
`f(a)=a`
`f(a+h)=a+h`
`f(a+2h)=a+2h`
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`f(a+bar(n-1)h)=a+bar(n-1)h`
`:.int_a^b x dx=lim_(h->0)h[a+(a+h)+` `(a+2h)+....+(a+bar(n-1)h)]`
`=lim_(h->0)h[a+a+.... n` times]`+ h(1+2+...+bar(n-1))]`
`=lim_(h->0)h(an+h((n-1)n/2)]`
`=lim_(h->0)[anh+((nh-h)(nh)/2]`
`=lim_(h->0)[a(b-a)+((b-a-h)(b-a))/2]`
`=a(b-a)+(b-a)^2/2`
`=ab-a^2+(b^2+a^2-2ab)/2`
`=(2ab-2a^2+b^2+a^2-2ab)/2`
`=(b^2-a^2)/2`
Question: 2. `int_0^5(x+1)dx`
Solution: `int_0^5(x+1)dx`
Here `f(x)=x+1, a=0,` `b=5, nh=5-0=5`
We know that, `int_a^b f(x)dx`
`=lim_(h->0) h[f(a)+f(a+h)+` `f(a+2h) ....+f(a+bar(n-1)h)]`
`f(a)=f(0)=0+1=1`
`f(a+h)=f(o)=h+1`
`f(a+2h)=f(2h)=2h+1`
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`f(a+bar(n-1)h)=f((bar(n-1)h)=bar(n-1)h+1`
`:. int_0^5(x+1)dx`
`=lim_(h->0)h[1+(h+1)+(2h+1)` `+....+bar(n-1)h+1]`
`=lim_(h->0)h[(1+1+1+....n` times)`+h(1+2+....+bar(n-1))]`
`=lim_(h->0)h[n+h.((n-1)n)/2]`
`=lim_(h->0)[nh+((nh-h)(nh))/2]`
`=lim_(h->0)[5+((5-h)(5))/2]`
[∵ `nh=5`]
`=5+((5-0)(5))/2`
`=5+25/2=35/2`
Question: 3. `int_2^3 x^2 dx`
Solution: Given, `int_2^3 x^2 dx`
We know that, `int_a^b f(x)dx`
`=lim_(h->0) h[f(a)+f(a+h)+` `f(a+2h) ....+f(a+bar(n-1)h)]`
`f(a)=f(2)=2^2=4`
`f(a+h)=f(2+h)=(2+h)^2` `=4+h^2+4h`
`f(a+2h)=f(2+2h)=(2+2h)^2` `=4+4h^2+8h`
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`f(a+bar(n-1)h)=f(2+bar(n-1)h)` `=(2+bar(n-1)h)^2` `=4+(n-1)^2h^2+4(n-1)h`
`:. int_2^3 x^2 dx`
`=lim_(h->0)h[4+(4+h^2+4h)+` `(4+4h^2+8h)+....`
`+(4+(n-1)^2h^2+4(n-1)h)]`
`=lim_(h->0)h[(4+4+4+....n` times)
`+h^2(1^2+4+....+(n-1)^2)`
`+4h(1+2+...+(n-1)))]`
`=lim_(h->0)h[4n+h^2[1^2+2^2+....`
`+(n-1)^2]+4h((n-1)n)/2]`
`=lim_(h->0)[4n+h^2((n-1)(n)(2n-1))/6` `+2h(n-1)*n]`
`lim_(h->0)[4nh+((nh-h)(nh)(2nh-h))/6` `2(nh-h)*nh]`
`=lim_(h->0)[4xx1+((1-h)(1)(2xx1-h))/6` `+2(1-h)(1)]`
`=lim_(h->0)[4+((1-h)(1)(2-h))/6` `+2(1-h)]`
`=4+((1-0)(1)(2-0))/6+2(1-0)`
`=4+2/6+2`
`=6+1/3+19/3`
Question: 4.`int_1^4(x^2-x)dx`
Solution: Given, `int_1^4(x^2-x)dx`
Here, `f(x)=x^2-x, a=1`,
`b=4, nh=4-1=3`
We know that, `int_a^b f(x)dx`
`=lim_(h->0) h[f(a)+f(a+h)+` `f(a+2h) ....+f(a+bar(n-1)h)]`
`f(a)=f(1)=1^2-1=0`
`f(a+h)=f(1+h)`
`=(1-h)^2-(1+h)`
`=1+h^2+2h-1-h=h^2+h`
`f(a+2h)=f(1+2h)`
`=(1+2h)^2-(1+2h)`
`=1+4h^2+4h-1-2h`
`=4h^2+2h`
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`f(a+bar(n-1)h)=f(1+bar(n-1)h)`
`=(1+bar(n-1)h)^2-(1-bar(n-1)h)`
`=1+(n-1)^2h^2+2(n-1)h`
`-1-(n-1)h`
`=(n-1)^2h^2+(n-1)h`
`:.int_1^4(x^2-x)dx`
`lim_(h->0)h[0+(h^2+h)+(4h^2+2h)`
`+....+(n-1)^2h^2+(n-1)h)]`
`=lim_(h->0)h(h^2(1+4+...+(n-1)^2`
`+h(1+2+...+(n-1))]`
`=lim_(h->0)h[h^2(1^2+2^2+...+(n-1)^2]`
`+h((n-1)n)/2`
`=lim_(h->0)[h^2((n-1)(n)(2n-1))/6`
`+(h(n-1)(n))/2]`
`=lim_(h->0)[((nh-h)(nh)(2nh-h))/6`
`+((nh-h)(nh))/2]`
`=lim_(h->0)[((3-h)(3)(2xx3-h))/6` `+((3-h)(3))/2]`
`=lim_(h->0)[((3-h)(3)(6-h))/6` `+((3-h)(3))/2]`
`=((3-0)(3)(6-0))/6+((3-0)(3))/2`
`=9+9/2=27/2`
Question: 5. `int_-1^1 e^x dx`
Solution: `int_-1^1 e^x dx`
Here, `f(x) = e^x, a=-1, b=1`
`nh=1-(-1)=1+1=2`
We know that, `int_a^b f(x)dx`
`=lim_(h->0) h[f(a)+f(a+h)+` `f(a+2h) ....+f(a+bar(n-1)h)]`
`f(a)=f(-1) = e^-1`
`f(a+h)=f(-1+h)=e^(-1+h)`
`f(a+2h)=f(-1+2h)=e^(-1+2h)`
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`f(a+bar(n-1)h)=f(-1+bar(n-1)h)` `=e^(-1+(n-1)h)`
`:. int_-1^1 e^xdx`
`=lim_(h->0)h[e^-1+e^(-1+h)+e^(-12h)` `+....+e^(-1+(n-1)h)]`
`=lim_(h->0)h(e^-1+e^-1*e^h+e^-1*e^2h` `+.....+e^-1*e^((n-1)h)]`
`=e^-1 lim_(h->0)h[1+e^h+e^2h` `+...+e^((n-1)h)]`
`=1/h lim_(h->0)h[((e^h)^n-1)/(e^h-1)]`
[∵ In a G.P., `S_n=(a(r^n-1))/(r-1)]`
`=1/e lim_(h->0)h[(e^(nh)-1)/(e^h-1)]`
`=1/e lim_(h->0)h[(e^2-1)/(e^h-1)]`
`=1/e lim_(h->0) h ((e^2-1))/((e^h-1)/h)`
`=1/e (e^2-1)=(e-1/e)`
Question:6. `int_0^4(x+e^(2x))dx`
Solution: `int_0^4(x+e^(2x))dx`
Here, `f(x)=x+e^(2x), a=0,` `b=4, nh=4-0=4`
We know that, `int_a^b f(x)dx`
`=lim_(h->0) h[f(a)+f(a+h)+` `f(a+2h) ....+f(a+bar(n-1)h)]`
`f(a)=f(0)` `=0+e^(2xx0)=e^0=1`
`f(a+h)=f(h)=f+e^(2h)`
`f(a+2h)=f(2h)=2h+e^(4h)`
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`f(a+bar(n-1)h)` `=f(bar(n-1)h)` `=(n-1)h+e^(2(n-1)h)`
`:. int_0^4(x+e^(2x))dx`
`lim_(h->0)h[1+(h+e^(2h))+(2h+e^(4h))` `+....+(n-1)h+e^(2(n-1)h)]`
`=lim_(h->0)h [h(1+2+...+(n-1)` `+(1+e^(2h)+e^(4h)+...+e^(2(n-1)h))]`
`=lim_(h->0)h[h((n-1)h)/2+((e^(2h)^n-1))/(e^(2h)-1)]`
`=lim_(h->0)[((nh-h)(nh))/2+((e^(2nh)-1))/((e^(2h)-1)/(2h))*2]`
`=lim(h->0)[((4-h)(4))/2+(e^8-1)/((e^(2n)-1)/(2h)*2)]`
`=((4-1)(4))/2+(e^8-1)/(1xx2)`
`=8+(e^8-1)/2`
`=(16+e^8-1)/2`
`=(e^8-15)/2`
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