Integrals NCERT Solutions
Solution of NCERT Exercise: 7.9: Q 15-22
Question: (15) `int_0^1 x e^(x^2) dx`
Solution:
Given, `int_0^1 x e^(x^2) dx`
Now, `int x e^(x^2) dx`
Put `x^2=t=>2xdx = dt`
`=> xdx = 1/2 dt`
`:. int x e^(x^2)dx=1/2 int e^t dt`
`=1/2 e^t = 1/2 e^(x^2)`
By second fundamental theorem, we get
`int_0^1 x\ e^(x^2) dx=[1/2 e^(x^2)]_0^1`
`=1/2 e^(1^2)-1/2e^0`
`=1/2e - 1/2 = 1/2(e-1)`
Question: (16)`int_1^2(5x^2)/(x^2+4x+3) dx`
Solution:
Given, `int_1^2(5x^2)/(x^2+4x+3) dx`
Now, `int (5x^2)/(x^2+4x+3)dx`
`=int[5-(5(4x+3))/(x^2+4x+3)]dx`
`=5int dx - 5int (4x+3)/((x+1)(x+3))`
`=5x-5I_1` ---(i)
Now, `I_1=int(4x+3)/((x+1)(x+3))dx`
The integral `(4x+3)/((x+1)(x+3))` is a proper rational function
`:.(4x+3)/((x+1)(x+3))=A/(x+1)+B/(x+3)` ---(ii)
`=>4x+3=A(x+3)+B(x+1)`
`=>4x+3=(A+B)x+(3A+B)`
Comparing coeficients of like terms on both sides, we have
`A+B=4` ----(iii)
`3A+B=3` ----(iv)
Subtracting (iii) from (iv), we get
`A=-1/2`
Putting value A in (iii), we get
`B=9/2`
Putting values of A and B in (ii) we get
`(4x+3)/((x+1)(x+3))=-1/(2(x+1))+9/(2(x+3))`
`:. int (4x+3)/((x+1)(x+3))dx`
`=int[-1/(2(x+1))+9/(2(x+3))]dx`
`=-1/2int1/(x+1)dx+9/2int 1/(x+3)dx`
`=-1/2log(x+1)+9/2log(x+3)`
Putting the value of I1 in (i), we get
`int(5x^2)/(x^2+4x+3)dx`
`=5x-x+5/2[-1/2log(x+1)+9/2log(x+3)]`
`5x+5/2log(x+1)-45/2log(x+3)`
By second fundamental theorem, we get
`int_1^2 (5x^2)/(x^2+4x+3)dx`
`=[5x+5/2log(x+1)-log(2+3)]` `-[5xx1+5/2log(1+1)-45/2log(1+3)]`
`=10+5/2log3-45/2log5-5` `-5/2log2+45/2log4`
`=5+5/2(log3-log2)` `-45/2(log5-log4)`
`=5+5/2log(3/2)-45/2log(5/4)`
Question: (17) `int_0^(pi//4)(2sec^2 x +x^3+2)dx`
Solution:
Given, `int_0^(pi//4)(2sec^2 x +x^3+2)dx`
Now, `int(2xdec^2x+x^3+2)dx`
`=2int sec^2x dx +int x^3 dx +2 int dx`
`=2tanx +x^4/4+2x`
By second fundamental theorem, we get
`int_0^(pi//4)(2sec^2 x +x^3+2)dx`
`=[2tanx +x^4/4+2x]_0^(pi//4)`
`=[2tan\ pi/4+(pi//4)^2/4+2xxpi/4]` `-[2tan(0)+(0)^4/4+2xx0]`
`=2+pi^4/1024+pi/2-0`
`=2+pi^4/1024+pi/2`
Question: (18) `int_0^pi(sin^2\ x/2-cos^2\ x/2)dx`
Solution:
Given, `int_0^pi(sin^2\ x/2-cos^2\ x/2)dx`
Now`int(sin^2\ x/2-cos^2\ x/2)dx`
`=-int(cos^2\ x/2-sin^2\ x/2)dx`
`=-int cos x dx`
`=-sinx`
By second fundamental theorem, we get
`int_0^pi(sin^2\ x/2-cos^2\ x/2)dx`
`=[-sinx ]_0^pi`
`=(-sin pi)-(sin0)`
`=0-0 = 0`
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