How do you find the definite integral of #x/sqrt(2x+1)# from #[0,4]#?

Answer 1

Use substitution to evaluate #int x/sqrt(2x+1) dx#. Then evaluate from #0# to #4#.

Let #int x/sqrt(2x+1) dx#
Let #u = 2x+1#. This makes #du = 2 dx#, so #dx = 1/2 du#
And it also makes #x = 1/2(u-1)#.
Furthermore, when #x=0#, we have #u=1# and
when #x = 4#, we get #u = 9#

Substitute:

#int x/sqrt(2x+1) dx = int (1/2(u-1))/(sqrtu) * 1/2 du = 1/4 int (sqrtu - 1/sqrtu) du#
# = 1/4[2/3 u^(3/2)-2u^(1/2)] = 1/6sqrtu(u-3)#
#int_0^4 x/sqrt(2x+1) dx = 1/4 int_1^9 (sqrtu - 1/sqrtu) du#
# = {: 1/6sqrtu(u-3)]_1^9 = 3/6(6) - (-2)/6 = 20/6 = 10/3#
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Answer 2

To find the definite integral of ( \frac{x}{\sqrt{2x+1}} ) from 0 to 4, you can use the substitution method. Let ( u = \sqrt{2x+1} ), then ( x = \frac{u^2 - 1}{2} ) and ( du = \frac{1}{\sqrt{2x+1}} dx ).

After substitution, the integral becomes ( \int_{1}^{3} \frac{u^2 - 1}{2u} du ). Simplify the integrand to ( \frac{1}{2} \int_{1}^{3} (u - \frac{1}{u}) du ).

Integrate term by term: ( \frac{1}{2} \left[ \frac{u^2}{2} - \ln|u| \right]_{1}^{3} ).

Substitute back for ( u ): ( \frac{1}{4} \left[ (3^2 - 1) - \ln(3) - (1^2 - 1) + \ln(1) \right] ).

Simplify: ( \frac{1}{4} \left[ 8 - \ln(3) \right] ).

So, the definite integral of ( \frac{x}{\sqrt{2x+1}} ) from 0 to 4 is ( \frac{1}{4} \left[ 8 - \ln(3) \right] ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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