Evaluate ∫(𝑥+3)√4−𝑥^2𝑑𝑥 by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.
sorry the integral is from x=-2 to x=2 so would the first integral equal to 0?
a=-2, b=2*
To evaluate the given integral, let's start by decomposing it into a sum of two integrals.
∫(𝑥+3)√(4−𝑥^2)𝑑𝑥
Let's set up the decomposition:
∫(𝑥+3)√(4−𝑥^2)𝑑𝑥 = ∫(𝑥√(4−𝑥^2))𝑑𝑥 + ∫(3√(4−𝑥^2))𝑑𝑥
Now, let's interpret one of these integrals in terms of an area.
The second integral, ∫(3√(4−𝑥^2))𝑑𝑥, represents the area under the curve y = 3√(4−𝑥^2) from x = -2 to x = 2.
In this integral, the function 3√(4−𝑥^2) represents the height of each infinitesimally thin rectangle, and 𝑑𝑥 represents the width. Integrating this expression will give us the total area between the curve and the x-axis within the given limits.
To find the value of ∫(𝑥+3)√(4−𝑥^2)𝑑𝑥, we can evaluate both integrals individually.
First, let's focus on the first integral:
∫(𝑥√(4−𝑥^2))𝑑𝑥
This integral can be solved using substitution. Let u = 4 - 𝑥^2:
𝑑𝑢/𝑑𝑥 = -2𝑥
𝑑𝑢 = -2𝑥𝑑𝑥
𝑑𝑥 = (1/-2)𝑑𝑢
Substituting u and 𝑑𝑥 into the integral:
∫(𝑥√(4−𝑥^2))𝑑𝑥 = ∫((𝑢-4)√𝑢)(1/-2)𝑑𝑢
= (-1/2)∫(𝑢√𝑢)𝑑𝑢 - (-1/2)∫(4√𝑢)𝑑𝑢
Now, let's evaluate each integral separately:
∫(𝑢√𝑢)𝑑𝑢 = (2/5)𝑢^(5/2) + 𝐶1
∫(4√𝑢)𝑑𝑢 = (8/3)𝑢^(3/2) + 𝐶2
Plugging in the values, we have:
(-1/2)∫(𝑢√𝑢)𝑑𝑢 - (-1/2)∫(4√𝑢)𝑑𝑢
= (-1/2)((2/5)𝑢^(5/2) + 𝐶1) - (-1/2)((8/3)𝑢^(3/2) + 𝐶2)
= (-1/5)𝑢^(5/2) + 𝐶1/2 + (4/3)𝑢^(3/2) - 𝐶2/2
Now, let's move on to the second integral:
∫(3√(4−𝑥^2))𝑑𝑥
We previously interpreted this integral as the area under the curve y = 3√(4−𝑥^2) from x = -2 to x = 2.
To find this area, we can apply the formula for the area of a quarter of a circle:
Area = (1/4)πr^2
In this case, the radius r is equal to 2, so the area is:
(1/4)π(2^2) = (1/4)π(4) = π/4
Therefore, interpreting the second integral in terms of an area gives us π/4.
Putting it all together, we have:
∫(𝑥+3)√(4−𝑥^2)𝑑𝑥 = (-1/5)𝑢^(5/2) + 𝐶1/2 + (4/3)𝑢^(3/2) - 𝐶2/2 + π/4
Remember to substitute u back in terms of x to get the final result.
∫(𝑥+3)√(4−𝑥^2) 𝑑𝑥 = ∫x √(4−𝑥^2) 𝑑𝑥 + 3∫√(4−𝑥^2) 𝑑𝑥
The first term is -1/2 the area under the parabola y=√(4-x^2)^3
The second is 3 times the area under a semicircle of radius 2.
no. you just have the full area under the curves.
Go ahead and do the integration!