Finding The Area Under The Curve Using Definite Integrals - Calculus
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- Опубликовано: 28 ноя 2023
- This calculus video tutorial explains how to find the area under the curve using definite integrals in terms of x and y.
Antiderivatives: • Antiderivatives
U-Substitution - Indefinite Integrals:
• How To Integrate Using...
1st Order Differential Equations:
• Separable First Order ...
Initial Value Problem:
• Initial Value Problem
Area Between Two Curves:
• Area Between Two Curves
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Disk and Washer Method:
• Disk & Washer Method -...
Volume By The Shell Method:
• Shell Method - Volume ...
Volume By Cross Sections:
• Volumes Using Cross Se...
Arc Length Calculus Problems:
• Arc Length Calculus Pr...
Surface Area of Revolution:
• Surface Area of Revolu...
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Work Problems - Calculus:
• Work Problems - Calculus
Integration By Parts:
• Integration By Parts
Trigonometric Integrals:
• Trigonometric Integrals
Trigonometric Substitution:
• Trigonometric Substitu...
Integration By Partial Fractions:
• Integration By Partial...
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Please check out Professor Leonard/ Professor V videos on the Calculus Sequence.
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This is a must-watch for students looking to conquer calculus challenges. The video's straightforward approach and practical examples demystify the process, making it easier for anyone to understand and apply definite integrals in finding the area under a curve.
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Professor Organic Chemistry Tutor, thank you for an exceptional video/lecture on Finding the Area Under a Curve using Definite Integrals in Calculus Two. From the video, the best way to find the Area under a Curve is to graph the function(s) and then calculate the required Area. Thanks to the viewers for finding and correcting the errors in this video.
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Can you do a video on LaPlace Transform please!
When we try to find the area under the graph for solution of x, as in x = y^2 for example, do we change it to y = √x so we solve for y? Because I went to Symbolab (site for mathematics) and it solved for y. But it also showed the graph as it is on the y axis even though it solved for y? Why do we solve for y in this case? Should not we solve for x?
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Hey could we please look at something like 5-3e ^-2t between 2 and 4 please i understand this = 1/a e (^at) +c but im a little stumped on how to work this out completely? thank you.
my calculus professor said that if the if the shaded area is under the x-axis, it contributes negatively to the integral so then for 11:26 the area would be 0 but you are saying this isn't the case? why?
You are correct. He accidentally did the proof for functions that are even (he even said BEFORE he started doing the work that the answer was 0).
In reality, if a function is odd like y = f(x), and you want to find the area between points -4 and 4, it will always be 0 since there is an equal amount of "negative area" contributed with "positive area".
HOWEVER, if the function is an even function, like y = x^2, and you want to find the area between -4 and 4 (equally opposite bounds), you can just find ONE side and multiply by 2. I think the Organic Chemist did the wrong proof on accident, but yes you are right, the problem at 11:26 is indeed equal to 0, not 16.
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Quick twist -
In equation -x^2 + 6x -8 how do we find the value of y that would yield area under the curve equals to 4. Currently area is calculated with lower Y bound equal to 0.
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What do we achieve when calculating for F(a) and F(b)?
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Find the curve bounded by the positive x,y axes and the curve y=4-×^2
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Calculate the area enclosed by y=1÷×^2-1 and y=3
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