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How do you evaluate the definite integral #int (8)/sqrt(3+4x)# from #[0,1]#?

Answer 1

definite integral #=4(sqrt7-sqrt3)#

Use the substitution #u=3+4x# Then #du/dx=4# #dx=(du)/4# So the integral is #int(8dx)/(sqrt(3+4x))=int(2du)/sqrtu=2*sqrtu/(1/2)=4sqrtu# going back to x #int_0^1(8dx)/(sqrt(3+4x))=4sqrt(3+4x)∣_0^1=4sqrt7-4sqrt3#

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Answer 2

Isabella Ackerman

To evaluate the definite integral (\int_{0}^{1} \frac{8}{\sqrt{3+4x}} , dx), we can use the substitution method.

Let (u = 3 + 4x). Then, (du = 4 , dx) and (dx = \frac{1}{4} du).

When (x = 0), (u = 3 + 4(0) = 3). When (x = 1), (u = 3 + 4(1) = 7).

Now, substituting into the integral:

[ \int_{0}^{1} \frac{8}{\sqrt{3+4x}} , dx = \int_{3}^{7} \frac{8}{\sqrt{u}} \cdot \frac{1}{4} , du ]

[ = 2 \int_{3}^{7} \frac{1}{\sqrt{u}} , du ]

[ = 2 \left[2\sqrt{u}\right]_{3}^{7} ]

[ = 4 \left(\sqrt{7} - \sqrt{3}\right) ]

So, (\int_{0}^{1} \frac{8}{\sqrt{3+4x}} , dx = 4(\sqrt{7} - \sqrt{3})).

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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.