How do you find the indefinite integral of #int 4/sqrt(5t)dt#?

Answer 1

THe answer is #=(8sqrtt)/sqrt5+C#

We need

#intx^ndx=x^(n+1)/(n+1) +C#
#int1/sqrtxdx=intx^(-1/2)dx=x^(1/2)/(1/2)=2sqrtx#

So,

#int(4dt)/sqrt(5t)=4/sqrt5*2sqrtt+C#
#=(8sqrtt)/sqrt5+C#
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Answer 2

To find the indefinite integral of ( \int \frac{4}{\sqrt{5t}} dt ), you can use the substitution method. Let ( u = \sqrt{5t} ). Then, ( du = \frac{1}{2\sqrt{5t}} dt ). Rearrange to solve for ( dt ), yielding ( dt = \frac{2}{\sqrt{5}} du ). Substitute these expressions into the integral:

[ \int \frac{4}{\sqrt{5t}} dt = \int \frac{4}{u} \cdot \frac{2}{\sqrt{5}} du ]

[ = \frac{8}{\sqrt{5}} \int \frac{1}{u} du ]

The integral ( \int \frac{1}{u} du ) integrates to ( \ln|u| + C ), where ( C ) is the constant of integration. Substituting back in for ( u ) yields:

[ \frac{8}{\sqrt{5}} \ln|\sqrt{5t}| + C ]

[ = \frac{8}{\sqrt{5}} \ln|(\sqrt{5})\sqrt{t}| + C ]

[ = \frac{8}{\sqrt{5}} \ln|(\sqrt{5t})| + C ]

Therefore, the indefinite integral of ( \int \frac{4}{\sqrt{5t}} dt ) is ( \frac{8}{\sqrt{5}} \ln|(\sqrt{5t})| + C ), where ( C ) is the constant of integration.

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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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