निम्नलिखित समाकलनो का मान ज्ञात कीजिए:
(i) `int_(2)^(3) x^(2) dx` (ii) `int_(4)^(9) sqrt(x)/((30-x .^(3)//_(2))^(2)) dx`
(iii) `int_(1)^(2) x/((x+1)(x+2)) dx` (iv) `int_(0)^(pi//4) sin^(3) 2t cos 2t dt`

Explanation

(i) `int_(2)^(3) x^(2) dx=[x^(3)/3]_(2)^(3)`
`=1/3 [(3)^(3) -(2)^(3)]=1/3 [27-8]`
`=19/3`.
(ii) `int_(4)^(9) sqrt(x)/((30-x^(3//2))^(2))`
मान लीजिए `t=30-x^(3//2)`
`dt=-3/2 x^(1//2) dx=-3/2 sqrt(x) dx`
या `sqrt(x) dx=-2/3 dt`
`:. int_(4)^(9) sqrt(x)/((30-x^(3//2))^(2)) dx =-2/3 int_(4)^(9) (dt)/t^(2)=-2/3 [(-1/t)]_(4)^(9) =2/3 [(1/(30-x^(3//2)))]_(4)^(9)`
`=2/3 [1/(30-9^(3/2))-1/(30-4^(3//2))]`
`=2/3 [1/(30-27)-1/(30-8)]=2/3 (1/3-1/22)=19/99`
(iii) `int_(1)^(2) x/((x+1)(x+2)) dx`
`:. (x)/((x+1)(x+2))=-1/(x+1) +2/(x+2)` [आंशिक भिन्न के प्रयोग से]
`:. int (x)/((x+1)(x+2)) dx =-int (dx)/(x+1)+2 int (dx)/(x+2)`
`=-log (x+1)+2 log (x+2)`
अब `int_(1)^(2) (x)/((x+1)(x+2)) dx=[-log (2+1)+2 log (2+2)]-[-log (1+1)+2 log (1+2)]`
`=-log 3+2 log 4 +log 2-2 log 3`
`=-3 log 3 + 2 log 2^(2) + log 2`
`=-3 log3+4 log 2 + log 2`
`=-3 log 3+5log 2`
`=log 2^(5) -log 3^(3)`
`=log 32-log 27`
`=log 32/27`
(iv) `int_(0)^(pi//4) sin^(3) 2t cos 2t dt`
मान लीजिए `sin 2t=x`
`2 cos 2t dt =dx`
`cos 2t dt =1/2 dx`
`:. int_(0)^(pi//4) sin^(3) 2t cos 2t dt=1/2 int_(0)^(pi//4) x^(3) dx`
`=1/8 [x^(4)]_(0)^(pi//4) =1/8 [sin^(4) 2t]_(0)^(pi//4)`
`=1/8[sin^(4) 2xxpi/4-sin^(4) 2xx0]`
`=1/8 ["sin"^(4) pi/2-sin^(4) 0]`
`=1/8`