Question Number 101439 by mhmd last updated on 02/Jul/20

Commented bymhmd last updated on 02/Jul/20

help me sir ?

Commented bybramlex last updated on 02/Jul/20

(Q3)(dy/(4x^2 )) −sec 5y dx = 0 (dy/(sec 5y)) = 4x^2 dx ⇒∫cos 5y dy = (4/3)x^3 +C (1/5)sin 5y = (4/3)x^3 + C ★

Commented bybemath last updated on 02/Jul/20

(q4) 8x^2 = 32 , x = ± 2 The area = 2∫_0 ^2 (32−8x^2 ) dx = 2[(32x−(8/3)x^3 )]_0 ^2 = 2 (64−((64)/3)) = 128(1−(1/3)) = 128×(2/3) = ((256)/3) ♠

Commented bymhmd last updated on 02/Jul/20

thank you sir

Commented bybemath last updated on 04/Jul/20

(q2) ∫_0 ^2 [( 4y+(1/4)sin 4y)]_0 ^( π) dx =∫_0 ^2 (4π) dx = [ 4πy ]_0 ^2 = 8π ★

Commented bymhmd last updated on 02/Jul/20

sir can you help me in question one

Commented bybobhans last updated on 02/Jul/20

(Q5) cos (−2x) = 1−(((−2x)^2 )/(2!)) + (((−2x)^4 )/(4!)) −(((−2x)^6 )/(6!)) +...

Commented bymhmd last updated on 02/Jul/20

sir the interval (0 to pi)

Commented bymhmd last updated on 02/Jul/20

can you help me in question one

Commented bymhmd last updated on 03/Jul/20

no sir the result (8pi)not (16pi)