aofsorular.com
MAT110U

Indefinite Integral

1. Ünite 20 Soru
S

What is antiderivative of (y/2) eyx+5 ?

(½) eyx+5 + c

S

What is antiderivative of (5/2) e5x+y

(½) e5x+y + c

S

What is antiderivative of 6x2 + 8x + (3/x) + 4 ?

2x3 + 4x2 + 3lnx + 4x + c

S

What is antiderivative of x1/3 ?

(¾) x4/3 + c

S

What is antiderivative of x1/2 with F(1) = 2 ?

(2/3) x3/2 + c ; F(1) = 2 = (2/3) 13/2 + c = (2/3) + c ; c = 2 – (2/3) = 4/3

S

What is antiderivative of x-1/2 with F(0) = -1 ?

2 x1/2 + c ; F(0) = -1 = 2 01/2 + c = 0 + c ; c = -1

S

∫ ((3/x) + x) (x+1) dx = ?

∫ (3 + (3/x) + x2 + x) dx = 3x + 3lnx + (1/3) x3 + (1/2) x2 + c

S

∫ ((1 / x2) + 2x) (x + 3x2) dx = ?

 ∫ ((1/x) + 3 + 2x2 + 6x3) dx] = lnx + 3x + (2/3) x3 + (3/2)x4 + c

S

∫ ((x + y - 2) / (x + y)1/2) dx = ?

Change of variable : u = x + y ; du = dx ;

∫ ((u – 2) / (u1/2)) du = ∫ ((u / (u1/2)) – (2 / (u1/2)) du = ∫ (u1/2 – (2 / (u1/2)) du

= (2/3) u3/2 - 4 u1/2 + c

= (2/3) (x + y)3/2 - 4 (x + y)1/2 + c

S

∫ ((yx + 2) / (yx + 1)1/2) dx = ?

Change of variable : u = xy + 1 ; (1/y) du = dx ;

∫ ((u + 1) / (u1/2)) du = ∫ [((u / (u1/2)) + (1 / (u1/2)) du = ∫ (u1/2 + (1 / (u1/2)) du

= (2/3) u3/2 + 2 u1/2 + c

= (2/3) (x + y)3/2 + 2 (x + y)1/2 + c

S

∫ x2 (x3 + 1)5 dx = ?

Change of variable : u = x3 + 1 ; (1/3) du = x2 dx ;

∫ (1/3) u5 du] = (1/18) u6 + c =

(1/18) (x3 + 1)6 + c

S

∫ x (yx2 + 2)3 dx = ?

Change of variable : u = yx2 + 2 ; (1/2y) du = x dx ;

∫ (1/2y) u3 du] = (1/8y) u4 + c =

(1/8y) (yx2 + 2)4 + c

S

∫ ((3x2 + 1) / (x3 + x + 1)) dx = ?

Change of variable : u = (x3 + x + 1) ; du = (3x2 + 1) dx ;

∫ (1/u) du = lnu + c

= ln(x3 + x + 1) + c

S

∫ (2x (2x2 + 1) / (x4 + x2 + 2)) dx = ?

Change of variable : u = (x4 + x2 + 2) ; du = (4x3 + 2x) dx = 2x (2x2 + 1) dx ;

∫ (1/u) du = lnu + c =

ln(x4 + x2 + 2) + c

S

∫ ((1/2) – x) e-x dx = ?

Integration by parts : A = ∫ [u dv = uv - ∫ v du = uv – B ;

u = (1/2) – x ; du = - dx ;

dv = e-x dx ; v = - e-x ;

uv = ((1/2) – x) (- e-x) ;

vdu = (- e-x) (- dx) = e-x dx ; B = - e-x + c ;

A = ((1/2) – x) (- e-x) - (- e-x) + c = ((1/2) – x – 1) (- e-x) + c

= ((1/2) + x) e-x + c

S

∫ (x/y) e2x dx = ?

Integration by parts : A = ∫ u dv = uv - ∫ v du = uv – B ;

u = x/y ; du = (1/y) dx ;

dv = e2x dx ; v = (1/2) e2x ;

uv = (x/y) (1/2) e2x = (x/2y) e2x ;

vdu = (1/2) e2x (1/y) dx = (1/2y) e2x dx ; B = (1/2y) (1/2) e2x + c ;

A = (x/2y) e2x - (1/2y) (1/2) e2x + c

= (1/2y) (x – (1/2)) e2x + c

S

∫ x2 ln(x/2) dx = ?

Integration by parts : A = ∫ u dv = uv - ∫ v du = uv – B ;

u = ln(x/2) ; du = (1/x) dx ;

dv = x2 dx ; v = (1/3) x3 ;

uv = ln(x/2) (1/3) x3 ;

vdu = (1/3) x3 (1/x) dx = (1/3) x2 dx ; B = (1/3) (1/3) x3 + c ;

A = ln(x/2) (1/3) x3 - (1/3) (1/3) x3 + c

= (1/3) x3 (ln(x/2) – (1/3)) + c

S

∫ -x ln(yx) dx = ?

Integration by parts : A = ∫ u dv = uv - ∫ v du = uv – B ;

u = ln(yx) ; du = (1/x) dx ;

dv = -x dx ; v = (-1/2) x2 ;

uv = ln(yx) (-1/2) x2 ;

vdu = (-1/2) x2 (1/x) dx = (-1/2) x dx ; B = (-1/2) (1/2) x2 + c ;

A = ln(yx) (-1/2) x2 - (-1/2) (1/2) x2 + c

= (-1/2) x2 (ln(yx) – (1/2)) + c

S

∫ 5yx ex/2 dx = ?

Integration by parts : A = ∫ u dv = uv - ∫ v du = uv – B ;

u = 5yx ; du = 5y dx ;

dv = ex/2 dx ; v = 2 ex/2 ;

uv = 5yx 2 ex/2 = 10yx ex/2 ;

vdu = 2 ex/2 5y dx = 10y ex/2 dx ; B = 20y ex/2 + c ;

A = 10yx ex/2 - 20y ex/2 + c

= 10y (x – 2) ex/2 + c

S

∫ (7yx + 9zx) e5x dx = ?

Integration by parts : A = ∫ u dv = uv - ∫ v du = uv – B ;

(7yx + 9zx) = (7y + 9z) x = ax

u = ax ; du = a dx ;

dv = e5x dx ; v = (1/5) e5x ;

uv = ax (1/5) e5x ;

vdu = (1/5) e5x a dx ; B = (a/5) (1/5) e5x + c ;

A = ax (1/5) e5x - (a/5) (1/5) e5x + c = (a/5) (x – (1/5)) e5x + c

= ((7y + 9z)/5) (x – (1/5)) e5x + c