How do you solve #int_0^1 sqrt(5x+4) dx#?

Answer 1

#38/15#

We need to use a u-substitution to solve this question.

STEP 1: Identify the u #u = 5x+4#
STEP 2: Find du #du = 5 dx#
STEP 3: Change the x-bounds to u-bounds #x = 0 -> u = 5(0)+4 = 4# #x = 1 -> u = 5(1)+4 = 9#
STEP 4: Do the u-substitution #int_0^1 sqrt 5x+4 dx# #=int_4^9 sqrt(u) * 1/5 du# remember: we found #du = 5dx#, so if we solve for #dx# we get #dx=1/5 du# #=1/5 int_4^9 sqrt(u) du# #=1/5 [2/3 u^(3/2)]_4^9# #=2/15[u^(3/2)]_4^9# #=2/15(3^3 - 2^3)# #=2/15(19)# #=38/15#
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Answer 2

To solve the integral ∫₀¹ √(5x + 4) dx, we can use the substitution method. Let u = 5x + 4. Then, du/dx = 5, or du = 5 dx. Solving for dx, we get dx = du/5.

Now, we substitute u = 5x + 4 and dx = du/5 into the integral:

∫₀¹ √(5x + 4) dx = ∫₄⁹ √u * (1/5) du

This simplifies to:

(1/5) ∫₄⁹ √u du

Now, we integrate √u with respect to u:

(1/5) * (2/3) * [u^(3/2)] from 4 to 9

Substituting the limits of integration:

(1/5) * (2/3) * [(9^(3/2)) - (4^(3/2))]

This further simplifies to:

(1/5) * (2/3) * [(27) - (8)]

And finally, evaluating:

(1/5) * (2/3) * [19]

= 38/15

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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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