A i ) ∫ [ x − 2 x 2 ] [ x + 2 x 2 ] d x = ∫ [ x 2 − 4 x 4 ] d x = x 3 3 + 4 3 x 3 + c i i ) ∫ e 5 x [ e 2 x 7 + 3 e 3 x ] d x = ∫ [ e 7 x 7 + 3 e 2 x ] d x = e 7 x 49 + 3 e 2 x 2 + c i i i ) ∫ 1 ( 4 − 3 x ) 3 d x put 4 − 3 x = t = > − 3 2 3 x d x = d t ⟹ d x = − 2 3 ( 4 − t ) d t ∫ 1 ( 4 − 3 x ) 3 d x = − 2 3 ∫ 4 − t t 3 d t = − 2 3 ∫ ( 4 t − 3 − t − 2 ) d t = − 2 3 ( − 2 t 2 + 1 3 t 3 ) + c = − 2 3 ( − 2 ( 4 − 3 x ) 2 + 1 3 ( 4 − 3 x ) 3 ) + c i v ) ∫ 0 π / 4 ( t a n x ) 3 ( s e c x ) 3 d x put , s e c x = t = > t a n x s e c x d x = d t and using t a n 2 x = 1 − s e c 2 x , we get ∫ 0 π / 4 ( t a n x ) 3 ( s e c x ) 3 d x = ∫ 0 π / 4 ( s e c x ) 2 ( t a n x ) 2 t a n x s e c x d x = ∫ 0 π / 4 ( s e c x ) 2 ( 1 − s e c 2 x ) t a n x s e c x d x = ∫ 1 2 t 2 ( 1 − t 2 ) d t = ∫ 1 2 ( t 2 − t 4 ) d t = [ t 3 3 − t 5 5 ] 1 2 = − 0.32 B i ) A\\
i)\\
\int [x-\frac{2}{x^2}][x+\frac{2}{x^2}]dx\\
=\int [x^2-\frac{4}{x^4}]dx\\
= \frac{x^3}{3}+\frac{4}{3x^3}+c\\
ii)\\
\int e^{5x}[\frac{e^{2x}}{7}+\frac{3}{e^{3x}}]dx\\
=\int [\frac{e^{7x}}{7}+3{e^{2x}}]dx\\
=\frac{e^{7x}}{49}+\frac{3e^{2x}}{2}+c\\
iii)\\
\int \frac{1}{(4-\sqrt{3x})^3}dx\\
\text{put}\\
4-\sqrt{3x}=t\\
=>\frac{-3}{2\sqrt{3x}}dx=dt\\
\implies dx=\frac{-2}{3}(4-t)dt\\
\int \frac{1}{(4-\sqrt{3x})^3}dx\\
=\frac{-2}{3} \int \frac{4-t}{t^3}dt\\
=\frac{-2}{3} \int (4t^{-3}-t^{-2})dt\\
=\frac{-2}{3}(\frac{-2}{t^{2}}+\frac{1}{3t^3})+c\\
=\frac{-2}{3}(\frac{-2}{(4-\sqrt{3x})
^{2}}+\frac{1}{3(4-\sqrt{3x})^3})+c\\
iv)\\
\int^{π/4}_{0 } (tan x)^3 (sec x )^3dx\\
\text{put},\\
secx=t\\
=>tanx \space secx \space
dx=dt\\
\text{and using }tan^2x=1-sec^2x,
\text{we get}\\
\int^{π/4}_{0 } (tan x)^3 (sec x )^3dx\\
=\int^{π/4}_{0 } (secx)^2(tan x )^2tanx \space secx\space dx\\
=\int^{π/4}_{0} (sec x )^2(1-sec^2x)tanx \space secx\space dx\\
=\int^{\sqrt2}_1 t^2(1-t^2)dt\\
=\int^{\sqrt2}_1 (t^2-t^4)dt\\
=[\frac{t^3}{3}-\frac{t^5}{5}]^{\sqrt2}_1\\
=-0.32\\
B\\
i) A i ) ∫ [ x − x 2 2 ] [ x + x 2 2 ] d x = ∫ [ x 2 − x 4 4 ] d x = 3 x 3 + 3 x 3 4 + c ii ) ∫ e 5 x [ 7 e 2 x + e 3 x 3 ] d x = ∫ [ 7 e 7 x + 3 e 2 x ] d x = 49 e 7 x + 2 3 e 2 x + c iii ) ∫ ( 4 − 3 x ) 3 1 d x put 4 − 3 x = t => 2 3 x − 3 d x = d t ⟹ d x = 3 − 2 ( 4 − t ) d t ∫ ( 4 − 3 x ) 3 1 d x = 3 − 2 ∫ t 3 4 − t d t = 3 − 2 ∫ ( 4 t − 3 − t − 2 ) d t = 3 − 2 ( t 2 − 2 + 3 t 3 1 ) + c = 3 − 2 ( ( 4 − 3 x ) 2 − 2 + 3 ( 4 − 3 x ) 3 1 ) + c i v ) ∫ 0 π /4 ( t an x ) 3 ( sec x ) 3 d x put , sec x = t => t an x sec x d x = d t and using t a n 2 x = 1 − se c 2 x , we get ∫ 0 π /4 ( t an x ) 3 ( sec x ) 3 d x = ∫ 0 π /4 ( sec x ) 2 ( t an x ) 2 t an x sec x d x = ∫ 0 π /4 ( sec x ) 2 ( 1 − se c 2 x ) t an x sec x d x = ∫ 1 2 t 2 ( 1 − t 2 ) d t = ∫ 1 2 ( t 2 − t 4 ) d t = [ 3 t 3 − 5 t 5 ] 1 2 = − 0.32 B i )
blue curve denotes f = x 2 − 2 f=x^2-2 f = x 2 − 2
red curve denotes g=-|x|
(ii)
area of region I i.e., ABC,
2 × 1 2 × 1 × 1 = 1 2×\frac{1}{2}×1×1\\
=1 2 × 2 1 × 1 × 1 = 1
area of region II i.e., ADC,
2 × ∫ − 1 1 − ( x 2 − 2 ) d x = 2 × ∫ − 1 1 ( 2 − x 2 ) d x = 2 [ 2 x − x 3 3 ] − 1 1 = 2 [ 2 − 1 3 + 2 − 1 3 ] = 20 3 2×\int_{-1}^1 -(x^2-2)dx\\
=2×\int_{-1}^1 (2-x^2)dx\\
=2[2x-\frac{x^3}{3}]_{-1}^1\\
=2[2-\frac{1}{3}+2-\frac{1}{3}]\\
=\frac{20}{3} 2 × ∫ − 1 1 − ( x 2 − 2 ) d x = 2 × ∫ − 1 1 ( 2 − x 2 ) d x = 2 [ 2 x − 3 x 3 ] − 1 1 = 2 [ 2 − 3 1 + 2 − 3 1 ] = 3 20
total area of the region bounded i.e., ABCD is 23/3.
