int_(-2)^2sqrt(4-x^2)text()dx evaluate the integral
To evaluate the given integral: ∫(-2)^2√(4-x^2) dx
Step 1: Simplify the expression inside the square root:
√(4 - x^2)
This is the equation of a circle with a radius of 2 centered at the origin. It represents the upper half of the circle. Therefore, for the given integral, we need to evaluate only the upper half of the circle.
Step 2: Visualize the problem:
Draw the graph of the upper half of the circle, which ranges from -2 to 2 on the x-axis.
Step 3: Set up the integral:
∫(-2)^2√(4-x^2) dx
= ∫4√(4-x^2) dx
Step 4: Identify the trigonometric substitution:
Since the equation is related to a circle, let's use a trigonometric substitution to simplify the integral. Let x = 2sinθ, where -π/2 ≤ θ ≤ π/2.
Step 5: Calculate dx:
Differentiate x = 2sinθ with respect to θ to find dx:
dx = 2cosθ dθ
Step 6: Substitute x and dx into the integral:
∫4√(4 - (2sinθ)^2) (2cosθ) dθ
Simplify further:
∫4√(4 - 4sin^2θ) (2cosθ) dθ
∫4√(4(1 - sin^2θ)) (2cosθ) dθ
∫4√(4cos^2θ) (2cosθ) dθ
∫4(2cosθ) (2cosθ) dθ
Step 7: Simplify the integral:
∫16cos^2θ dθ
Now we have a standard integral that we can evaluate.
Step 8: Use the trigonometric identity:
cos^2θ = (1 + cos2θ)/2
∫(16(1 + cos2θ)/2) dθ
∫8(1 + cos2θ) dθ
Step 9: Evaluate the integral:
Let's integrate each term separately:
∫8 dθ + ∫8cos2θ dθ
The integral of a constant is:
8θ + ∫8cos2θ dθ
Now we need to calculate the integral of cos2θ.
Step 10: Evaluate the integral of cos2θ:
Using the formula: ∫cos2θ dθ = (1/2)θ + (1/4)sin2θ + C
Where C is the constant of integration.
Using this formula, we can integrate ∫8cos2θ dθ:
= 8θ + (8/2)θ + (8/4)sin2θ + C
= 8θ + 4θ + 4sin2θ + C
= 12θ + 4sin2θ + C
Step 11: Final step - substitute back:
Since we used a trigonometric substitution, we need to substitute back θ for x:
= 12•arcsin(x/2) + 4sin(2•arcsin(x/2)) + C
And this is the final evaluated integral for ∫(-2)^2√(4-x^2) dx.