How do I evaluate #int_1^4sqrt(t) \(5 + 2 t) dt#?

Answer 1

I would expand the argument as:

#int_1^4sqrt(t)*5+sqrt(t)*2tdt=#

I can now separate the integrals, as:

#int_1^4sqrt(t)*5dt+int_1^4sqrt(t)*2tdt=#
Remembering that #sqrt(t)=t^(1/2)# and the rules to manipulate exponentes, to get:
#=int_1^4t^(1/2)*5dt+int_1^4t^(1/2+1)*2dt=# #=int_1^4t^(1/2)*5dt+int_1^4t^(3/2)*2dt=#
Taking the constant out of the integrals and remembering that the integral of #x^n# is #x^(n+1)/(n+1)#;
#5t^(1/2+1)/(1/2+1)+2t^(3/2+1)/(3/2+1)|_1^4# #10/3t^(3/2)+4/5t^(5/2)|_1^4=48,13#
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Answer 2

To evaluate ( \int_{1}^{4} \sqrt{t}(5 + 2t) , dt ), you would first distribute the square root term into the polynomial, then integrate each term separately. The integral would yield ( \frac{32}{3}\sqrt{4} - \frac{4}{3}\sqrt{1} + \frac{5}{3}(4^{\frac{3}{2}} - 1) ).

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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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