emergency!! helpp!! (1 Viewer)

[totallybord]

hi!
someone please help me with :
1) what is the integral of e^x /(5 + e^x) ?
2) what is the volume of x^2/4 + y^2 = 4 ? rotated about the x axis... i did the limits from 0 to 4 and got the wrong answer...
3) what is the integral of 1/ (5x^5) ?
thank you heaps!

 

[SoulSearcher]

1 - Note the derivative of the denominator, i.e. the derivative of 5 + e^x, then compare that result to the numerator and note that the indefinite integral of f'(x)/f(x) w.r.t.x (where f(x) is any function) is equal to ln [f(x)] + C.

2 - The equation describes the graph of an ellipse, and therefore to find the volume in this particular case you must integrate with the limits at the respective endpoints of the graph, in this case -4 and 4.

3 - Note that the indefintie integral of xⁿ w.r.t.x is equal to xⁿ⁺¹/(n+1) and the relation that x^-n = 1/(xⁿ)

 

[kurt.physics]

Hi--
i havent done calculus for 2 months, but i think i've got it right, please all critiziation wanted.

Q1. No idea

Q2. First re-arrange equation in terms of x

x² / 4 + y² = 4

y² + 4 / x² = 1/4 (flip)

y² + 4 = x² / 4

y² = x² / 4 - 4

Now,

V = / π * ∫ y² * dx / (between -4 and 0) + π * ∫ y² * dx (between 0 and 4) note: // is absolute value
= π( 1/4 * ∫x² *dx -4 * ∫ dx ) note: * is multiply

= π( 1/4 * 1/3 * x³ -4x)

= π( x³ / 12 -4x)[between 0 and -4]

= π( 0 - -4³ / 12 -4 *-4)

= π(0 - -64/12 + 16)

= / π( 0 - -32/3) /

= / (32π/3) /

= 32π/3

∫ y² * dx = as we know is
= x³ / 12 -4x

(between 4 and 0) = π * (4³ / 12 -4*4 - 0)
= / (-32π/3) /

= 32π/3

Volume =/ π * ∫ y² * dx / (between -4 and 0) + π * ∫ y² * dx (between 0 and 4)

= 32π/3 + 32π/3

= 64π/3
= 67.02 (2.d.p)


Q3) ∫ 1 / 5x⁵ dx

1 / 5x⁵ = 1/5 * x^-5

∫ 1 / 5x⁵ dx = 1/5 ∫ x^-5 dx

= 1/5 * x^-4 / -4

= 1/5 * 1/-4 * 1/ x⁴

= 1/-20 * 1/ x⁴

= 1 / -20x⁴


Thats it, please, correct me if im wrong