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### Course: Integral Calculus>Unit 3

Lesson 4: Area: vertical area between curves

# Worked example: area between curves

The area between the graphs of functions ƒ and 𝑔 can be found by taking the definite integral of ƒ-𝑔 (suppose ƒ is above 𝑔).

## Want to join the conversation?

• At , why is the area positive if most of the graph is below the x-axis?
• Area is always positive, the negative values of the integrals make it positive although the if you're finding only the value of integral, the negative is accepted.
• After he says that when finding area b/w curves....it makes sense to subtract one equation from another and then integrate. He also said that there are many videos in which he has discussed that why it makes sense. Could someone pls tell me which videos is he talking about
• How can we find the lower and upper bounds without referring to the graph?
• If you have equations solved for the same variable, you can set them equal to each other in order to find the points on intersections. These points are your bounds.
• how can you tell top vs. bottom
• You don't need to know which one's top or bottom. Take the negative signs out of the integral if you want. Here, the net area is always positive. So, you can try taking the absolute value of the area if it's negative. Both are okay!
Hope it helps.
• Could you take the anti-derivative of the two functions in the beginning and then subtract them?
• Is it possible to find the area between two curves without looking at their graphs?
• If you can solve for where the functions intersect each other and determine which function is greater than the other in each interval between intersections, then you would not need to look at their graphs before integrating.
• I got the same answer taking the difference between the definite integral of the "upper" function and the "lower" function (definite integral of X^4 +4x^2+1 minus definite integral of x^2-3)

I don't understand why the first definite integral give me a negative value (precisely, -4/15). I expected a positive value, representing the area between the x axys and the blue function
• Could someone please explain what to do if you have two equations, y=2lnx and y=x-3, and how to get the coordinate points needed?
(1 vote)
• Generally, an equation with one side linear and the other side logarithmic cannot be solved algebraically for x. So it's best to use a graphing calculator or equation solver to find the x-coordinate(s) of the intersections of the graphs of y=2lnx and y=x-3.
• Can somebody help me, please? I'm trying to solve the below equation for x, and I'm on a step that's giving me confusion. When I see step 1, it's clear that one of the solutions for x is -7, but that solution applies no longer when I move on to step 2. What's happening??

Step 1: (x+7) = 0.25(x+7)^2
step 2: 1 = 0.25(x+7) ## divided both sides by the same amount, (x+7).
(1 vote)
• The reason your equation is breaking is that you're dividing by 0. x+7=0 when x=-7.