Integrals NCERT Solutions
NCERT Exercise 7.9
Evaluate the definite integrals in Exercises 1 to 20.
Question: 1. `int_-1^1 (x+1)dx`
Solution: `int_-1^1 (x+1)dx`
Now, `int (x+1)dx`
`=int xdx +int dx`
`= x^2/2+x`
By second fundamental theorem, we get
`int_-1^1 (x+1)dx =[x^2/2+x]_-1^1`
`=[(-1)^2/2+1]-[(-1)^2/2-1]`
`=3/2+1/2=2`
Question: 2. `int_2^3 1/xdx`
Solution: `int_2^3 1/xdx`
Now, `int 1/xdx = logx`
By second fundamental theorem, we get
`int_2^3 1/xdx =[logx]_2^3`
`=log3-log2=log\ 3/2`
Question: 3. `int_1^3(4x^3-5x^2+6x+9)dx`
Solution:
`int_1^3(4x^3-5x^2+6x+9)dx`
Now, `int(4x^3-5x^2+6x+9)dx``
`=4intx^3dx-5intx^2dx` `+6intxdx+9intdx`
`=(4x^4)/4-(5x^3)/3+(6x^2)/2+9x`
`=x^4-3/3x^3+3x^2+9x`
By second fundamental theorem, we get
`int_1^3(4x^3-5x^2+6x+9)dx`
`=[x^4-5/3x^3+3x^2+9x]_1^2`
`=[(2)^4-2/3(2)^3+3(2)^2+9xx2]` `-[(1)^4-5/3(1)^3+3(1)^2+9xx1]`
`=[16-40/3+12+18]` `-[1-5/3+3+9]`
`=[40-40/3]-[13-5/3]`
`=98/3-34/3=64/3`
Question: 4. `int_0^(pi//4) sin 2x dx`
Solution:
Given, `int_0^(pi//4) sin 2x dx`
Now, `int sin 2x dx = -(cos2x)/2`
By second fundamental theorem, we get
`int_0^(pi//4) sin 2x dx`
`=[-(cos 2x)/2]_0^(pi//4)`
`=[-(cos2 xx pi/4)/2]-[-(cos2xx0)/2]`
`=0+1/2=1/2`
Question: 5. `int_0^(pi//2)cos 2x dx`
Solution:
Given, `int_0^(pi//2)cos 2x dx`
Now, `int cos 2x dx = (sin2x)/2`
By second fundamental theorem, we get,
`int_0^(pi//2)cos 2x dx`
`=[(sin2x)/2]_0^(pi//2)`
`=[(sin2xxpi/2)/2]-[(sin2xx0)/2]`
`=1/2sin pi - 1/2 sin(0)=0`
Question: 6. `int_0^4 e^x dx`
Solution: `int_0^4 e^x dx`
Now, `int e^x dx = e^x`
By second fundamental theorem, we get
`int_0^4 e^x dx = [e^x]_4^5`
`=e^5-e^4=e^4(e-1)`
Question: 7. `int_0^(pi//4) tan x dx`
Solution: `int_0^(pi//4) tan x dx`
Now, `int tan x dx = log|secx|`
By second fundamental theorem, we get
`int_0^(pi//4) tan x dx`
`=[log|secx|]_0^(pi//4)`
`=log|sec\ pi/4|-log|sec\ 0|`
`=log(sqrt2)-log1`
`=1/2 log 2`
Question: (8) `int_(pi/6)^(pi/6) cosec\ x dx`
Solution:
Given, `int_(pi/6)^(pi/6) cosec\ x dx`
Now, `int cosec\ x dx =log|cosec\ x - cotx|`
By second fundamental theorem, we get
`int_(pi/6)^(pi/6) cosec\ x dx` `=[log|cosec\ x - cot x|]_(pi//6)^(pi//4)`
`=log|cosec\ pi/4 - cot\ pi/4|` `-log|cosec\ pi/6 - cot\ pi/6|`
`=log|sqrt2 - 1|-log|2-sqrt3|`
`=log|(sqrt2-1)/(2-sqrt3)|`
Question: (9) `int_0^1 (dx)/sqrt(1-x^2)`
Solution:
Given, `int_0^1 (dx)/sqrt(1-x^2)`
Now, `int(dx)/sqrt(1-x^2)=sin^-1\ x`
By second fundamental theorem, we get
`int_0^1 (dx)/sqrt(1-x^2)`
`=[sin^-1x]_x^1`
`=sin^-1(1)-sin^-1(0)`
`=pi/2-0=pi/2`
Question: (10) `int_0^1 (dx)/(1+x^2)`
Solution: Given, `int_0^1 (dx)/(1+x^2)`
Now, `1/(1+x^2)dx=tan^-1x`
By second fundamental theorem, we get
`int_0^1 1/(1+x^2)dx=[tan^-1x]_0^1`
`=tan^-1(1)-tan^-1(0)`
`=pi/4-0 = pi/4`
Question: (11) `int_2^3 (dx)/(x^2-1)`
Solution:
Given, `int_2^3 (dx)/(x^2-1)`
Now, `(dx)/(x^2-1)=int(dx)/((x)^2-(1)^2)`
`=1(2xx1)log|(x-1)/(x+1)|` `=1/2 log|(x-1)/(x+1)|`
By second fundamental theorem, we get
`int_2^3 (dx)/(x^2-1)` `=[1/2 lof|(x-1)/(x+1)]_2^3`
`=1/2 log|(3-1)/(3+1)|-1/2log|(2-1)/(2+1)|`
`=1/2[log1/2-log1/3]`
`=1/2 log (1//2)/(1//3)`
`=1/2 log 3/2`
Question: (12) `int_0^(pi//2)cos^2 x dx`
Solution: `int_0^(pi//2)cos^2 x dx`
Now, `int cos^2 x dx` `= int(1+cos2x)/2 dx`
`=1/2 int dx +1/2 int cos 2x dx`
`=x/2 +1/2 (sin2x)/2`
`= x/2 + (sin2x)/4`
By second fundamental theorem, we get
`int_0^(pi//2) cos^2 x dx=[x/2+(sin2x]/4]_0^(pi//2)`
`=pi/4+ (sin 2 xx pi/2)/4 -[0/2+(sin2xx0)/4]`
`=(pi/4+0)-0 = pi/4`
Question: (13) `int_2^3 x/(x^2+1)dx`
Solution: `int_2^3 x/(x^2+1)dx`
Now, `int x/(x^2+1)dx`
Put, `x^2+1=t`
`=>2xdx = dt`
`=>xdx = 1/2 dt`
`:. intx/(x^2+1)dx =1/2 int dt/t`
`=1/2 log t`
`=1/2 log(x^2+1)`
By second fundamental theorem, we get
`int_2^3 x/(x^2+1)dx` `= [1/2log(x^2+1]_2^3`
`=1/2 log(3^2+1)-1/2log(2^2+1)`
`=1/2log10-1/2log5`
`=1/2log(10/5)=1/2log2`
Question: (14) `int _0^1 (2x+3)/(5x^2+1)dx`
Solution:
Given, `int _0^1 (2x+3)/(5x^2+1)dx`
Now, `int(2x+3)/(5x^2+1)dx`
`=int (2x)/(5x^2+1)dx+3int 1/(5x^2+1)dx`
`=int (2x)/(5x^2+1)dx +3/5int 1/(x^2+1/5)dx`
`=I_1+3/5int1/(x^2+1/(sqrt5)^2)dx`
`=I_1+3/5xx 1/(1//sqrt5)tan^-1(x/(1//sqrt5))`
`=I_1+3/5 xx tan^-1(sqrt5 x)` ---(i)
Now, `I_1=int(2x)/(5x^2+1)dx`
Put, `5x^2+1 = t`
`=>10xdx = dt`
`=>2xdx = 1/5dt`
`:. int (2x)/(5x^2+1)dx = 1/5int (dt)/t`
`= 1/5log t = 1/5log(5x^2+1)`
Putting value of I1 in (i), we get
`int (2x+3)/(5x^2+1)dx`
`=1/5 log (5x^2+1)+3/sqrt5 tan^-1 (sqrt5 x)`
By second fundamental theorem, we get
`int_0^1 (2x+3)/(5x^2+1)dx`
`=[1/5log(5x^2+1)+3/sqrt5 tan^-1 (sqrt5x)]_0^1`
`=[1/5 log(5xx1^2+1)+3/sqrt5 tan^-1(sqrt5xx1)]` `-[1/5 log(5xx0+1)+3/sqrt5 tan^-1(sqrt5xx0)]`
`=[1/5 log(6)+3/sqrt5 tan^-1(sqrt5)]` `-1/5 log(1)+3/sqrt5 xx0]`
`=1/5 log (6) +3/sqrt5 tan^-1(sqrt5)-0`
`=1/5 log(6) + 3/sqrt5 tan^-1sqrt5`
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