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# Please solve this for me..

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Guest Dec 31, 2014

#1
+90088
+10

Here's (c)

5∫ x / √(x + 1) dx  .....let's integrate by "parts".....we'll go back and account for the "5" later on...

Let u = x   so...    du = dx

Let dv = 1 / √(x + 1)      and ..... v = 2 √(x + 1)   .... so we have

∫ u dv = u*v - ∫ v dv =  2x √(x + 1) - ∫ 2 √(x + 1) dx = 2 [x √(x + 1) - ∫  √(x + 1) dx ]

∫ u dv  =  2[x √(x + 1) - ∫  √(x + 1) dx ]

And, substituting in the limits for for the first term we have

(2)[3 √(3 + 1) - 1√(1 + 1)] = (2)[6 - √(2) ]

Now, for the second integral, let u = x + 1    du  = dx ... so we have....

-2∫ u^(1/2) du =

(-2)(2/3) u^(3/2)  =

(-4/3)(x + 1)^(3/2)

And substituting in the limits of integration, we have

(-4/3)[(3 + 1)^(3/2) - (1 + 1)^(3/2)] =

(4/3) [2^(3/2) - 4^(3/2)] =

(4/3)[ √8 - 8]

Putting everything together, we have

(2)(6 - √2 )  +  (4/3) (√8 - 8 ) = about 2.2761423749153967

But remenber, in the original integrlal, we have 5 times this much ....so we have about ..... 11.3807118745769835

CPhill  Jan 1, 2015
#1
+90088
+10

Here's (c)

5∫ x / √(x + 1) dx  .....let's integrate by "parts".....we'll go back and account for the "5" later on...

Let u = x   so...    du = dx

Let dv = 1 / √(x + 1)      and ..... v = 2 √(x + 1)   .... so we have

∫ u dv = u*v - ∫ v dv =  2x √(x + 1) - ∫ 2 √(x + 1) dx = 2 [x √(x + 1) - ∫  √(x + 1) dx ]

∫ u dv  =  2[x √(x + 1) - ∫  √(x + 1) dx ]

And, substituting in the limits for for the first term we have

(2)[3 √(3 + 1) - 1√(1 + 1)] = (2)[6 - √(2) ]

Now, for the second integral, let u = x + 1    du  = dx ... so we have....

-2∫ u^(1/2) du =

(-2)(2/3) u^(3/2)  =

(-4/3)(x + 1)^(3/2)

And substituting in the limits of integration, we have

(-4/3)[(3 + 1)^(3/2) - (1 + 1)^(3/2)] =

(4/3) [2^(3/2) - 4^(3/2)] =

(4/3)[ √8 - 8]

Putting everything together, we have

(2)(6 - √2 )  +  (4/3) (√8 - 8 ) = about 2.2761423749153967

But remenber, in the original integrlal, we have 5 times this much ....so we have about ..... 11.3807118745769835

CPhill  Jan 1, 2015