ज्ञात करे[ Evalute]` int _(0)^(3//2) | x cos pi x | dx `

ज्ञात करे[ Evalute]` int _(0)^(3//2) | x cos pi x

1.	ज्ञात करे[ Evalute]` int _(0)^(3//2) \| x cos pi x \| dx `
Answer» ` x cos pi x = 0 rArr {{:(x =0 ),(cos pix = 0 " या " pi x = (2n+1)pi/2 " , " n in Z ):}` ` rArr {{:( x=0 ),(x=1/2","0 " तथा " 3/2 " के बीच ") :}` अब , ` int _(0)^(3//2) \| x cos pi x \| dx = int _(0)^(1//2) \| x cos pix\| dx + int _(1//2)^(3//2) \| x cos pix \| dx ` ` = int _(0)^(1//2) x cos pi x dx - int _(1//2)^(3//2) x cos pix //dx` ` = [ (x sin pi x)/pi + (cos pix)/pi ]_(0)^(1//2) - [ (xsin pix)/pi+ (cos pix)/(pi^(2))]_(1//2)^(3//2)` ` = (1/(2pi)-1/(pi^(2)))-(-3/(2pi)-1/(2pi))= 1/(2pi) - 1/(pi^(2)) + 2/pi = 5/(2pi) - 1/(pi^(2))`

ज्ञात करे [Evaluate ] ` int_(0)^(pi//2) (x)/(sin x +cos x) dx `
निम्नलिखित को ज्ञात करे : (ii) `int_(0)^(2)xsqrt(2-x)dx`
ज्ञात करे [ Evaluate] `int_(1)^(4)f(x)dx, "` जहाँ (where) ` f(x) = |x -1| + |+| x-2| + | x- 3|`
ज्ञात करे ` int _(-1)^(1)x^(3) e^(x^(4))dx`
` int _(1)^(3)x^(3)dx` का मान ज्ञात कीजिए ।
निम्नलिखित को ज्ञात करे । (v) ` int _(-1)^(1) (|x|+|x -1|) dx`
माना कि ` I = int _(-pi)^(pi) (2x (1+ sin x))/(1+cos^(2) x)dx `
ज्ञात करे `int _(-pi//4)^(pi//4) x^(3) sin^(4) x dx `
ज्ञात करे[ Evalute]` int _(0)^(3//2) | x cos pi x | dx `
निम्नलिखित को ज्ञात करे [ Evalute the following ] : (iii)` int _(0)^(1)x(1-x)^(n) dx `