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# some simple integration problems that i cant seem to do...! watch

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1. integrate with respect to x
a) 4/√x + 3/x³
b) 6x(1+x²)^½

show that the substitution x=u² transforms ∫(limits 4 to 1) (1+√x)³/√x dx into an integral of the form ∫ (limits b to a) k(1+u)³du

that the value ok k, a and b. evaluate this integral
2. integrate with respect to x
a) 4/√x + 3/x³
b) 6x(1+x²)^½

show that the substitution x=u² transforms ∫(limits 4 to 1) (1+√x)³/√x dx into an integral of the form ∫ (limits b to a) k(1+u)³du

that the value ok k, a and b. evaluate this integral
3. (Original post by idiosyncratic88)
integrate with respect to x
a) 4/√x + 3/x³
b) 6x(1+x²)^½

show that the substitution x=u² transforms ∫(limits 4 to 1) (1+√x)³/√x dx into an integral of the form ∫ (limits b to a) k(1+u)³du

that the value ok k, a and b. evaluate this integral
a)

∫4/√x + 3/x³ dx = ∫4x^-0.5 + 3x^-3 dx = 8x^0.5 - 1.5x^-2 + c

b)

∫ 6x(1+x²)^½ dx = 2∫ 3x(1+x²)^½ dx = 2(1+x^2)^1.5 + c

c)

∫(1+√x)³/√x dx

k=u^2

∫(1+u)³/u du = ∫1/u + 3 + 3u + u^2 du =
4. (Original post by idiosyncratic88)
integrate with respect to x
a) 4/√x + 3/x³
b) 6x(1+x²)^½

show that the substitution x=u² transforms ∫(limits 4 to 1) (1+√x)³/√x dx into an integral of the form ∫ (limits b to a) k(1+u)³du

that the value ok k, a and b. evaluate this integral
a)

∫4/√x + 3/x³ dx = ∫4x^-0.5 + 3x^-3 dx = 8x^0.5 - 1.5x^-2 + c

b)

∫ 6x(1+x²)^½ dx = 2∫ 3x(1+x²)^½ dx = 2(1+x^2)^1.5 + c

c)

∫(1+√x)³/√x dx

k=u^2

∫(1+u)³/u du = ∫1/u + 3 + 3u + u^2 du =
5. a) ∫(4/√x + 3/x³)dx

= ∫(4x-1/2 + 3x-3)dx

= 8x1/2 - 3x-2/2 + c

I think ...

Hope this helps,

~~Simba
6. a) ∫(4/√x + 3/x³)dx

= ∫(4x-1/2 + 3x-3)dx

= 8x1/2 - 3x-2/2 + c

I think ...

Hope this helps,

~~Simba
7. ooh thanks!!! ill probably have loads more integrations Qu's that i'll need help with coming up!! if u like 'em ive got loads to give ya!
thanks again
8. ooh thanks!!! ill probably have loads more integrations Qu's that i'll need help with coming up!! if u like 'em ive got loads to give ya!
thanks again

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Updated: December 6, 2005
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