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Calculus 2 HELP PLEASE!!!?

0 votes
CAN YOU PLEASE SHOW EVERY STEP

1. FIND Y

y'=5sqrt(5x-1); y(1/5)=5

2. INTERGRATE
(1/X^2)(5^1/X) DX

3. D/DX (5^X + 1/SQRT(X) + piX + pi)

4 FIND DY/DX if Y=X^ln(x)

5. FIND DY/DX IF Y= INTERGAL X^2 AT THE TOP AND 1 AT THE BOTTOM
SQRT(1+T^3) DT

6. INTERGRATE
INTERGAL ln(9) at the top and ln(4) at the bottom e^X/2 DX

7. INTERGRATE
INTERGAL 1/3 e^2X COS(e^2x+1) DX

8. INTERGRATE
INTERGAL 2Xsec(X^2)DX

9. INTERGRATE
INTERGAL 2 at the top ad 1 at the bottom log base 2 (x)/x DX

10 INTERGRATE
V^2+V^1/2/V^3 DV
asked Apr 29, 2013 in CALCULUS by chrisgirl Apprentice

11 Answers

0 votes

ʃy' = ʃ5(√5x - 1)dx

5x - 1 = t2

Diferenciate each side with respective x

5dx = 2tdt

Substitute 5dx = 2tdt and 5x - 1 = t2

           = ʃt 2tdt

           = 2ʃt2dt

           = 2(t3 / 3) + C

Substitute t = √(5x - 1)

           = 2(√(5x - 1)3) + C

 

 

answered May 1, 2013 by diane Scholar
0 votes

 ʃ(1 / x2)51 / xdx

Let 1 / x  = t

Diferenciate each side with respective x

(-1 / x2)dx = dt

Substitute (-1 / x2)dx = dt and  1 / x = t

           = ʃ-dt5t

           = -ʃ5tdt

           = -(5t / log5) + C

Substitute t = 1 / x

           = -(51 / x / log5) + C.

 

 

answered May 1, 2013 by diane Scholar
0 votes

3.

d /dx (5x + 1 / sqrt(x) + πx + π)

Recall :  Derivative of (ax ) = ax loga and d / dx(1 / √(x)) = -1 / 2x√(x)

                                                        = 5x log5 + -1 / 2x√(x) + π + 0

                                                        = 5x log5 -1 / (2x√(x)) + π.

 

answered May 1, 2013 by diane Scholar
0 votes

4.

y = xlogx

Diferenciate each side with respective x

dy / dx = logx xlogx - 1 (1 /  x)

            = logx xlogx - 1( 1 /  x)

           = logx / x xlogx - 1 .

          

 

answered May 1, 2013 by diane Scholar
0 votes

ʃ1 / 3 e2x cos(e2x + 1 )dx  = 1 / 3ʃ e2xcos(e2x × e)dx

Let e2x = t

Diferenciate  each side with respective x

2e2xdx = dt

e2x dx = dt  / 2

Substitute e2x  = t and e2x dx = dt  / 2

                                            = 1 / 3ʃdt / 2 cos(te)

                                            = 1 / 6ʃcos(et) dt

                                            = 1 / 6(sinet) / e

                                           = 1 / 6e(sinet).

answered May 1, 2013 by diane Scholar
0 votes

7.

ʃ1 / 3 e2x cos(e2x + 1 )dx  = 1 / 3ʃ e2xcos(e2x × e)dx

Let e2x = t

Diferenciate  each side with respective x

2e2xdx = dt

e2x dx = dt  / 2

Substitute e2x  = t and e2x dx = dt  / 2

                                            = 1 / 3ʃdt / 2 cos(te)

                                            = 1 / 6ʃcos(et) dt

                                            = 1 / 6(sinet) / e

                                           = 1 / 6e(sinet).

answered May 1, 2013 by diane Scholar
0 votes

8.

ʃ2xsec(x2)dx = ʃ2xdxsec(x2)

Let x2 = t

Diferenciate  each side with respective x

2xdx = dt

Substitute x2 = t and 2xdx = dt

                                            = ʃdt sec(t)

                                            = ʃsec(t) dt

                                            = log|sec(t) + tan(t)| + C.

answered May 2, 2013 by diane Scholar
0 votes

10.

ʃ(V2 + V1 / 2  /  V3)DV = ʃ(V2 DV+ ʃ(V1 / 2 /  V3)DV

Integral V2  = V3 / 3 and integral Vn = Vn + 1

                                           = V3 / 3 +ʃ V -5 / 2DV

                                          = V3 / 3 + V -3 / 2 / -3 / 2 + C

                                          = V3 / 3 +(-2 / 3) V -3 / 2 + C.

answered May 2, 2013 by diane Scholar
0 votes

4) y = x^iogx

dy/dx = d/dx(x^logx)

let logx = t

Then differentiate each side

dt = 1/x*dx

dy/dx = d/dt(x^t)

dy/dx = t*x^(t-1)dt   { d/dx(x^n) = n*x^(n-1)}

dy/dx = tx^(t-1)(1/x)

substitute t = logx

dy/dx = logx*x^(logx - 1)(1/x)

dy/dx =( logx*x^(logx - 1) )/x

answered May 14, 2013 by jeevitha Novice
0 votes

6) image

By Substitution method

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Now the integral becomes

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answered Jul 14, 2014 by david Expert

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