Definite Integral - Area Bounded Between Circle X^2+Y^2=4 And (X-2)^2+Y^2=4

Definite Integral - Area bounded between circle x^2+y^2=4 and (x-2)^2+y^2=4

Let's find the area between the two circles using definite integrals.
First, let's find the intersection points of the two circles.
The equations of the circles are:
- Circle 1: \(x^2 + y^2 = 4\)
- Circle 2: \((x-2)^2 + y^2 = 4\)
By solving these equations simultaneously, we get:
- Intersection point 1: (0, 2)
- Intersection point 2: (2, 0)
Now, let's find the area between these two intersection points.

The area between two curves can be found using the formula: \[A = \int_{a}^{b} (f(x) - g(x)) \, dx\]
In this case, the curves are the circles, so we need to find the functions \(f(x)\) and \(g(x)\).
For the upper curve, we can take \(f(x) = \sqrt{4 - x^2}\).
Similarly, for the lower curve, we can take \(g(x) = -\sqrt{4 - (x-2)^2}\).
Now, let's substitute these values in the area formula.

Substituting the values of the functions, the area formula becomes: \[A = \int_{0}^{2} (\sqrt{4 - x^2} + \sqrt{4 - (x-2)^2}) \, dx\]
We can simplify the equation by expanding the square roots and combining like terms.
After simplification, the equation becomes: \[A = \int_{0}^{2} (2\sqrt{x^2 - x + 1}) \, dx\]
Let's move on to evaluating this definite integral.

To evaluate the definite integral, we can use the substitution method.
Let's substitute \(x^2 - x + 1\) as \(t^2\).
By differentiating both sides, we get \(2x - 1 = 2t \cdot \frac{dt}{dx}\).
Simplifying, we have \(dx = \frac{2t}{2x-1} \, dt\).
Now, substitute \(dx\) and \(x\) in the integral equation.

Substituting the values, the definite integral becomes: \[A = \int_{1}^{3} \left(2\sqrt{t^2} \cdot \frac{2t}{2x-1}\right) \, dt\]
Simplifying, we have: \[A = \int_{1}^{3} \frac{4t^2}{2t - 1} \, dt\]
Now, let's evaluate this definite integral.

To evaluate this integral, we can use the method of partial fractions.
Let's start by writing the fraction with the numerator of higher degree as the sum of two fractions.
After simplification, we get: \[\frac{4t^2}{2t-1} = 2t + \frac{2t}{2t-1}\]
Now, we can easily integrate both terms separately.
After integrating, we get the area as a result.

After evaluating the integrals, we get: \[A = \left[ t^2 + 2\ln|2t-1| \right]_{1}^{3}\]
Substituting the upper and lower limits, the area becomes: \[A = \left( 3^2 + 2\ln|5| \right) - \left( 1^2 + 2\ln|1| \right)\]
Further simplification gives us the final result.

Finally, after calculating, we get the area bounded between the given circles as:
\(A = 9 + 2\ln(5) - 1 - 0\)
\(A = 10 + 2\ln(5)\)

Sure! Here are slides 11 to 20 for the topic "Definite Integral - Area bounded between circle x^2+y^2=4 and (x-2)^2+y^2=4":

Let's recap what we have learned so far:
We started by finding the intersection points of the two circles.
Then, we defined the upper and lower curves as functions \(f(x)\) and \(g(x)\).
We used the area formula \[A = \int_a^b (f(x) - g(x)) \, dx\] to find the area between the curves.
We simplified the equation and applied substitution to evaluate the definite integral.

Now, let's move on to a different example to further understand definite integrals.
Example 1: Find the area bounded between the x-axis and the curve \(y = x^2 - 2x\) from x = 1 to x = 4.
We can start by sketching the graph of the given curve.
Next, we need to find the limits of integration (a and b) for the definite integral.
In this case, a = 1 and b = 4, as given in the question.

Example 1 (continued):
To find the area, we need to calculate the definite integral: \[A = \int_1^4 (x^2 - 2x) \, dx\]
We can simplify the equation by expanding \((x^2 - 2x)\).
After simplification, the equation becomes: \[A = \int_1^4 (x^2 - 2x) \, dx\]
Let's evaluate this definite integral.

Example 1 (continued):
To integrate the function \((x^2 - 2x)\), we can split it into two separate integrals.
We have: \[A = \int_1^4 (x^2 \, dx) - \int_1^4 (2x \, dx)\]
We can now integrate both terms separately.
After integrating, we substitute the upper and lower limits to find the area.

Example 1 (continued):
Integrating the first term, we get: \[\int_1^4 (x^2 \, dx) = \left[ \frac{x^3}{3} \right]_1^4\]
Substituting the upper and lower limits, the first term becomes: \[\left( \frac{4^3}{3} \right) - \left( \frac{1^3}{3} \right)\]
Similarly, we can integrate the second term to find the area.
Finally, we subtract the two results to obtain the area.

Next, let's discuss the properties of definite integrals.
Property 1: \(\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\)
This property implies that reversing the limits of integration changes the sign of the definite integral.
Property 2: \(\int_a^b (f(x) + g(x)) \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx\)
This property states that the integral of the sum of two functions is equal to the sum of their individual integrals.

Property 3: \(\int_a^b kf(x) \, dx = k \int_a^b f(x) \, dx\)
This property allows us to factor a constant out of the definite integral.
Property 4: \(\int_a^a f(x) \, dx = 0\)
This property states that the integral of a function over the same limit is always zero.
These properties help us simplify and evaluate definite integrals.

Finally, let's revise the applications of definite integrals.
Application 1: Area between curves - We can use definite integrals to find the area between two curves.
Application 2: Volume of solids of revolution - We can use definite integrals to find the volume of solids obtained by rotating a curve around an axis.
Application 3: Work done - Definite integrals can be used to find the work done in moving an object against a force.
Application 4: Center of mass - Definite integrals help us find the center of mass of an object with varying density.

Sure! Here are slides 21 to 30 on the topic “Definite Integrals: Properties and Applications”:

Now, let's discuss some more properties of definite integrals.
Property 5: \(\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx\)
This property states that we can split the definite integral into two parts and the sum of the integrals will give the integral of the whole interval.
Property 6: \(\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx\)
This property allows us to split the definite integral into two smaller intervals.

Property 7: \(\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^d f(x) \, dx + \int_d^b f(x) \, dx\)
This property extends property 6 to more than two smaller intervals.
Property 8: \(\int_a^b f(x) \, dx + \int_b^a f(x) \, dx = 0\)
This property states that if the limits of integration are reversed, the sum of the integrals is zero.
These properties help us manipulate definite integrals and simplify calculations.

Next, let's move on to the applications of definite integrals.
Application 5: Finding the length of a curve - Definite integrals can be used to find the length of a curve in the coordinate plane.
Application 6: Probability - Definite integrals can be used to calculate probabilities in various situations.
Application 7: Fluid pressure - Definite integrals help us determine the pressure exerted by a fluid on a given surface.
Application 8: Electric charge - Definite integrals are used to calculate the electric charge enclosed within a closed surface.

Application 9: Fourier series - Definite integrals play a crucial role in representing periodic functions using Fourier series.
Application 10: Complex analysis - Definite integrals are extensively used in the field of complex analysis to evaluate complex-valued functions.
Application 11: Heat transfer - Definite integrals help us determine the amount of heat transferred through a given surface.
Application 12: Optimal control - Definite integrals are used in optimal control theory to find the optimal path for a system.
These applications demonstrate the wide range of areas where definite integrals are applied.

Let's summarize what we have learned in this lecture:
Definite integrals are used to find the exact value of the area between curves, the volume of solids, and other important quantities.
They can be evaluated using various methods including substitution, integration by parts, and partial fractions.
Definite integrals have properties that help us manipulate and solve complex problems.
They have diverse applications in mathematics, physics, engineering, and other fields.

Now, let's solve a few practice problems to reinforce our understanding of definite integrals.
Practice Problem 1: Find the area bounded between the x-axis and the curve \(y = \sin(x)\) from x = 0 to x = \(\pi\).
Practice Problem 2: Calculate the volume of the solid obtained by rotating the curve \(y = x^2\) around the x-axis from x = 0 to x = 2.
Practice Problem 3: Determine the work done in moving an object against a force given by \(F(x) = 2x\) for x = 1 to x = 5.
Practice Problem 4: Find the center of mass of an object with density \(\rho(x) = e^x\) for x = 0 to x = \(\ln(2)\).

That brings us to the end of this lecture on definite integrals.
We have covered the definition, properties, and applications of definite integrals.
Remember to practice solving problems related to definite integrals to solidify your understanding.
If you have any doubts or questions, feel free to ask during the next class or office hours.
Thank you for your attention!