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show steps for finding the area of a function from points (10.11356,0) to (29.11...
Aug 13, 2024
show steps for finding the area of a function from points (10.11356,0) to (29.1114,0) using the function -40.1782 + 9.53583x - 0.836984x^2 + 0.0332897x^3 - 0.000486481x^4 | dx
Generated Graph
Solution by Steps
step 1
To find the area under the curve from x=10.11356 x = 10.11356 to x=29.1114 x = 29.1114 for the function −40.1782+9.53583x−0.836984x2+0.0332897x3−0.000486481x4 -40.1782 + 9.53583x - 0.836984x^2 + 0.0332897x^3 - 0.000486481x^4 , we need to compute the definite integral:
∫10.1135629.1114(−40.1782+9.53583x−0.836984x2+0.0332897x3−0.000486481x4) dx \int_{10.11356}^{29.1114} (-40.1782 + 9.53583x - 0.836984x^2 + 0.0332897x^3 - 0.000486481x^4) \, dx
step 2
Evaluating the integral gives us:
∫10.1135629.1114(−40.1782+9.53583x−0.836984x2+0.0332897x3−0.000486481x4) dx=61.3956 \int_{10.11356}^{29.1114} (-40.1782 + 9.53583x - 0.836984x^2 + 0.0332897x^3 - 0.000486481x^4) \, dx = 61.3956
step 3
The left sum approximation for the integral can be expressed as:
61.3956+40.1297n4−101.525n2−0.000115232n=61.3956−0.000115232n+O((1n)2) 61.3956 + \frac{40.1297}{n^4} - \frac{101.525}{n^2} - \frac{0.000115232}{n} = 61.3956 - \frac{0.000115232}{n} + O\left(\left(\frac{1}{n}\right)^2\right)
step 4
The antiderivative of the function is given by:
−0.0000972962x5+0.00832242x4−0.278995x3+4.76792x2−40.1782x+C -0.0000972962 x^5 + 0.00832242 x^4 - 0.278995 x^3 + 4.76792 x^2 - 40.1782 x + C
Answer
The area under the curve from x=10.11356 x = 10.11356 to x=29.1114 x = 29.1114 is approximately 61.3956 61.3956 .
Key Concept
The definite integral represents the area under a curve between two points on the x-axis.
Explanation
The calculated area of 61.3956 61.3956 indicates the net area between the curve and the x-axis over the specified interval.
show steps for finding the area of a function from points (0,0) to (29,0) using the function -0.030881 + 1.5074x - 0.10237x^2 + 0.00397811x^3 - 0.000079285x^4 | dx
Generated Graph
Solution by Steps
step 1
To find the area under the curve from x=0 x = 0 to x=29 x = 29 for the function −0.030881+1.5074x−0.10237x2+0.00397811x3−0.000079285x4 -0.030881 + 1.5074x - 0.10237x^2 + 0.00397811x^3 - 0.000079285x^4 , we need to compute the definite integral:
step 2
Evaluating the integral gives us: ∫029(−0.030881+1.5074x−0.10237x2+0.00397811x3−0.000079285x4) dx=178.897 \int_0^{29} (-0.030881 + 1.5074x - 0.10237x^2 + 0.00397811x^3 - 0.000079285x^4) \, dx = 178.897
step 3
The left sum approximation for the integral can be expressed as: 178.897+54.2075n4−254.782n2+20.7817n 178.897 + \frac{54.2075}{n^4} - \frac{254.782}{n^2} + \frac{20.7817}{n} (assuming subintervals of equal length)
step 4
The antiderivative of the function is: −0.000015857x5+0.000994528x4−0.0341233x3+0.7537x2−0.030881x+C -0.000015857x^5 + 0.000994528x^4 - 0.0341233x^3 + 0.7537x^2 - 0.030881x + C
Answer
The area under the curve from x=0 x = 0 to x=29 x = 29 is approximately 178.897 178.897
Key Concept
The area under a curve can be found using definite integrals.
Explanation
The definite integral calculates the total area between the curve and the x-axis over a specified interval, providing the net area.
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