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Math  /  Calculus

QuestionPrevious Problem Problem List Next Problem (1 point) Use part II of the Fundamental Theorem of Calculus evaluate the definite integral: ∫4182xdx=\int_{4}^{18} 2 \sqrt{x} d x= □\square
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Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of 2x2\sqrt{x} from x=4x = 4 to x=18x = 18. Watch out! Don't forget to apply the power rule correctly and be careful with your arithmetic when plugging in the limits of integration.

STEP 2

1. Rewrite the integrand
2. Find the indefinite integral
3. Evaluate the definite integral

STEP 3

Let's **rewrite** the integrand 2x2\sqrt{x} in a form that's easier to integrate.
Remember that the square root of xx is the same as xx raised to the power of one-half, so we can write 2x2\sqrt{x} as 2x122x^{\frac{1}{2}}.
This makes it easier to apply the power rule for integration!

STEP 4

Now, let's **find** the *indefinite integral* of 2x122x^{\frac{1}{2}}.
Using the power rule, we add 1 to the exponent and divide by the new exponent.
So, we have: ∫2x12dx=2⋅x12+112+1+C\int 2x^{\frac{1}{2}} dx = 2 \cdot \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C

STEP 5

Simplifying the exponent, 12+1=12+22=32\frac{1}{2} + 1 = \frac{1}{2} + \frac{2}{2} = \frac{3}{2}.
Substituting this back into our integral, we get: ∫2x12dx=2⋅x3232+C\int 2x^{\frac{1}{2}} dx = 2 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C

STEP 6

To divide by a fraction, we multiply by its reciprocal.
So, dividing by 32\frac{3}{2} is the same as multiplying by 23\frac{2}{3}: ∫2x12dx=2⋅23x32+C\int 2x^{\frac{1}{2}} dx = 2 \cdot \frac{2}{3}x^{\frac{3}{2}} + C

STEP 7

Multiplying the constants, we get our indefinite integral: ∫2x12dx=43x32+C\int 2x^{\frac{1}{2}} dx = \frac{4}{3}x^{\frac{3}{2}} + C

STEP 8

Now comes the exciting part!
We'll **evaluate** the *definite integral* from x=4x = 4 to x=18x = 18.
We'll use the indefinite integral we just found and substitute the **upper limit** of integration (1818) and the **lower limit** of integration (44): ∫4182xdx=[43x32]418=(43(18)32)−(43(4)32)\int_{4}^{18} 2\sqrt{x} dx = \left[ \frac{4}{3}x^{\frac{3}{2}} \right]_{4}^{18} = \left( \frac{4}{3}(18)^{\frac{3}{2}} \right) - \left( \frac{4}{3}(4)^{\frac{3}{2}} \right)

STEP 9

Let's **calculate** those values!
Remember that x32x^{\frac{3}{2}} is the same as x3\sqrt{x^3}: 43(18)32=43183=435832=43⋅722=962\frac{4}{3}(18)^{\frac{3}{2}} = \frac{4}{3}\sqrt{18^3} = \frac{4}{3}\sqrt{5832} = \frac{4}{3} \cdot 72\sqrt{2} = 96\sqrt{2} 43(4)32=4343=4364=43⋅8=323\frac{4}{3}(4)^{\frac{3}{2}} = \frac{4}{3}\sqrt{4^3} = \frac{4}{3}\sqrt{64} = \frac{4}{3} \cdot 8 = \frac{32}{3}

STEP 10

Finally, **subtract** the two results: 962−323=2882−32396\sqrt{2} - \frac{32}{3} = \frac{288\sqrt{2} - 32}{3}

STEP 11

Our **final answer** is 2882−323\frac{288\sqrt{2} - 32}{3}!

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