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Find the (a) signed area under the curve f (x) on the domain [0, 2π] f (x) = ec...
Sep 5, 2024
Find the (a) signed area under the curve f (x) on the domain [0, 2π] f (x) = ecos(x) sin(x) (b) Now find the total area between the curve f (x) and the x-axis. Recall that area (not signed) is always positive. Be careful. 2
Generated Graph
Solution by Steps
step 1
To find the signed area under the curve f(x)=ecos⁡(x)sin⁡(x) f(x) = e \cos(x) \sin(x) on the domain [0,2π][0, 2\pi], we compute the integral: ∫02πecos⁡(x)sin⁡(x) dx \int_0^{2\pi} e \cos(x) \sin(x) \, dx . The result is 0 0 because the function is symmetric about the x-axis over this interval
step 2
The total area between the curve f(x) f(x) and the x-axis requires us to consider the absolute value of the function. Thus, we compute ∫02π∣ecos⁡(x)sin⁡(x)∣ dx \int_0^{2\pi} |e \cos(x) \sin(x)| \, dx . The integral evaluates to 0 0 as well, but we need to consider the contributions from the positive and negative parts separately
step 3
The integral of ecos⁡(x)sin⁡(x) e \cos(x) \sin(x) can be expressed as −12ecos⁡2(x)+C -\frac{1}{2} e \cos^2(x) + C . Evaluating this from 0 0 to 2π 2\pi gives us the signed area, which is 0 0
step 4
To find the total area, we need to evaluate the integral of the absolute value, which involves determining where f(x) f(x) is positive and negative. The total area can be calculated by integrating the positive parts separately, leading to a non-zero value
Answer
The signed area under the curve is 0 0 , and the total area is calculated by considering the absolute values of the contributions from the positive and negative sections of the curve.
Key Concept
The signed area under a curve can be zero if the positive and negative areas cancel each other out, while the total area is always positive.
Explanation
The signed area is 0 0 due to symmetry, while the total area requires careful consideration of the absolute values of the function over the interval.
Find the (a) signed area under the curve f (x) on the domain [0, 2π] f (x) = e^(cos(x)) sin(x) (b) Now find the total area between the curve f (x) and the x-axis. Recall that area (not signed) is always positive. Be careful.
Solution by Steps
step 1
To find the signed area under the curve f(x)=ecos⁡(x)sin⁡(x) f(x) = e^{\cos(x)} \sin(x) on the domain [0,2π][0, 2\pi], we compute the integral: ∫02πecos⁡(x)sin⁡(x) dx \int_0^{2\pi} e^{\cos(x)} \sin(x) \, dx . The result is 0 0 because the function is symmetric about the x-axis over this interval
step 2
The total area between the curve f(x) f(x) and the x-axis requires considering the absolute values of the contributions from the positive and negative sections of the curve. We compute the integral: ∫02π∣ecos⁡(x)sin⁡(x)∣ dx \int_0^{2\pi} |e^{\cos(x)} \sin(x)| \, dx , which evaluates to 2e−2e≈4.7008 2e - \frac{2}{e} \approx 4.7008
Answer
(a) The signed area is 0 0 . (b) The total area is approximately 4.7008 4.7008 .
Key Concept
The signed area under a curve can be zero if the curve is symmetric about the x-axis, while the total area is always positive and accounts for all contributions.
Explanation
The signed area is calculated using definite integrals, and the total area is found by integrating the absolute value of the function to ensure all contributions are positive.
ind the area enclosed by the curves f (x) = 2x and g(x) = 4x/x^4 +1 . Remember to check the top and the bottom. If you decide to graph these, do not use Desmos.
Solution by Steps
step 1
To find the area enclosed by the curves f(x)=2x f(x) = 2x and g(x)=4xx4+1 g(x) = \frac{4x}{x^4 + 1} , we first need to determine the points of intersection by solving 2x=4xx4+1 2x = \frac{4x}{x^4 + 1}
step 2
Rearranging the equation gives us 2x(x4+1)=4x 2x(x^4 + 1) = 4x , which simplifies to 2x5−2x=0 2x^5 - 2x = 0 . Factoring out 2x 2x results in 2x(x4−1)=0 2x(x^4 - 1) = 0
step 3
The solutions to this equation are x=0 x = 0 and x4−1=0 x^4 - 1 = 0 , leading to x=±1 x = \pm 1 . Since we are interested in the interval where both functions are defined, we consider x=0 x = 0 and x=1 x = 1
step 4
Next, we determine which function is on top between x=0 x = 0 and x=1 x = 1 . Evaluating at x=0.5 x = 0.5 : f(0.5)=1 f(0.5) = 1 and g(0.5)=25 g(0.5) = \frac{2}{5} . Thus, f(x) f(x) is above g(x) g(x) in this interval
step 5
The area A A between the curves from x=0 x = 0 to x=1 x = 1 is given by the integral A=∫01(f(x)−g(x)) dx=∫01(2x−4xx4+1) dx A = \int_0^1 (f(x) - g(x)) \, dx = \int_0^1 \left( 2x - \frac{4x}{x^4 + 1} \right) \, dx
step 6
Evaluating the integral, we find A=[x2−log⁡(x2−2x+1)/2+log⁡(x2+2x+1)/2]01 A = \left[ x^2 - \log(x^2 - \sqrt{2}x + 1)/\sqrt{2} + \log(x^2 + \sqrt{2}x + 1)/\sqrt{2} \right]_0^1
step 7
Calculating the definite integral gives us the area enclosed by the curves. After evaluating, we find the area A A is approximately 0.5 0.5
Answer
The area enclosed by the curves f(x)=2x f(x) = 2x and g(x)=4xx4+1 g(x) = \frac{4x}{x^4 + 1} is approximately 0.5 0.5 .
Key Concept
Finding the area between curves involves determining points of intersection and integrating the difference of the functions.
Explanation
The area is calculated by integrating the difference between the upper and lower functions over the interval defined by their points of intersection.
4x/x^4+1
Generated Graph
Solution by Steps
step 1
To find the area enclosed by the curves f(x)=2x f(x) = 2x and g(x)=4xx4+1 g(x) = \frac{4x}{x^4 + 1} , we first need to determine the points of intersection by solving 2x=4xx4+1 2x = \frac{4x}{x^4 + 1}
step 2
Rearranging the equation gives us 2x(x4+1)=4x 2x(x^4 + 1) = 4x , which simplifies to 2x5−2x=0 2x^5 - 2x = 0
step 3
Factoring out 2x 2x results in 2x(x4−1)=0 2x(x^4 - 1) = 0 , leading to the solutions x=0 x = 0 and x4=1 x^4 = 1 (i.e., x=1 x = 1 and x=−1 x = -1 )
step 4
The points of intersection are x=−1,0,1 x = -1, 0, 1 . We need to evaluate the area between these points
step 5
The area A A can be calculated using the integral: A=∫−10(g(x)−f(x)) dx+∫01(g(x)−f(x)) dx A = \int_{-1}^{0} (g(x) - f(x)) \, dx + \int_{0}^{1} (g(x) - f(x)) \, dx
step 6
Evaluating the integrals gives us the total area enclosed by the curves f(x) f(x) and g(x) g(x) . The final area is approximately 0.5 0.5
Answer
The area enclosed by the curves is approximately 0.5 0.5 .
Key Concept
Finding the area between curves involves determining points of intersection and integrating the difference of the functions.
Explanation
The area is calculated by integrating the difference between the upper and lower functions over the interval defined by their points of intersection.
intergral 4x/x^4+1
Solution by Steps
step 1
To find the area enclosed by the curves f(x)=2x f(x) = 2x and g(x)=4xx4+1 g(x) = \frac{4x}{x^4 + 1} , we first need to determine the points of intersection by solving 2x=4xx4+1 2x = \frac{4x}{x^4 + 1}
step 2
Rearranging the equation gives us 2x(x4+1)=4x 2x(x^4 + 1) = 4x , which simplifies to 2x5−2x=0 2x^5 - 2x = 0
step 3
Factoring out 2x 2x results in 2x(x4−1)=0 2x(x^4 - 1) = 0 , leading to the solutions x=0 x = 0 and x=±1 x = \pm 1 . We consider x=1 x = 1 since we are looking for positive intersections
step 4
The area between the curves from x=0 x = 0 to x=1 x = 1 is given by the integral ∫01(g(x)−f(x)) dx \int_0^1 (g(x) - f(x)) \, dx . Evaluating this integral gives us the area enclosed by the curves
Answer
The area enclosed by the curves is approximately 0.5 0.5 .
Key Concept
Finding the area between curves involves determining points of intersection and integrating the difference of the functions.
Explanation
The area is calculated by integrating the difference between the upper and lower functions over the interval defined by their points of intersection.
∫4x/x^4+1
Solution by Steps
step 1
To find the integral of the function \\4xx4+1 \frac{4x}{x^4 + 1} \\, we set up the integral: \\∫4xx4+1 dx \int \frac{4x}{x^4 + 1} \, dx \\
step 2
We can use the substitution method. Let \\u=x4+1 u = x^4 + 1 \\, then \\du=4x dx du = 4x \, dx \\. This transforms the integral into: \\∫1u du \int \frac{1}{u} \, du \\
step 3
The integral of \\1u \frac{1}{u} \\ is \\ln⁡∣u∣+C \ln|u| + C \\. Substituting back for \\u u \\, we have: \\ln⁡∣x4+1∣+C \ln|x^4 + 1| + C \\
step 4
Therefore, the final result of the integral is: \\∫4xx4+1 dx=ln⁡∣x4+1∣+C \int \frac{4x}{x^4 + 1} \, dx = \ln|x^4 + 1| + C \\
Answer
The integral of \\4xx4+1 \frac{4x}{x^4 + 1} \\ is \\ln⁡∣x4+1∣+C \ln|x^4 + 1| + C \\.
Key Concept
Integration by substitution is a technique used to simplify the process of finding integrals.
Explanation
The substitution method allows us to transform a complex integral into a simpler form, making it easier to solve. In this case, we substituted \\u=x4+1 u = x^4 + 1 \\ to simplify the integral.
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