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The Learner's Stage
Aug 11, 2019

Will you please explain it?

We know that Differentiation and Integration of e^x is e^x. Doesn't it mean that the gradient and area covered under the graph of y = e^x is same. If so it is, then how?

6 comments
0
Lewis Cheung
Aug 15, 2019

Differentiation is a method to find the gradient of the curve.

Integration is a method to find the area under the curve from (minus infinity) to x.

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Marco Picerni
Aug 17, 2019

Yeah man, the area covered by e^x from 0 to A is e^A, and that value is exactly the gradient in that point, you can prove it by solving the limit h->0 of [f(x+h)-f(x)]/h, which describes the gradient of the graph in each point

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Phinehas
Aug 29, 2019

You mean negative infinity to A

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Ian Fowler
Nov 11, 2020  ·  Edited: Nov 11, 2020

The derivative is the slope of the tangent. i.e. the slope of the curve at a particular point. The gradient is the normal to the curve at the point of tangency and is therefore perp to the tangent. So the answer is no. At (a,e^a) the slope of the tangent is e^a. Therefore the slope of the gradient is -1/e^a


The area under e^x from -inf to A is e^a -- as stated by Phinehas.

The slope of the tangent to e^x at x = A is e^A


"Yeah man, the area covered by e^x from 0 to A is e^A"

No. The area under the curve e^x from 0 to A is (e^A - 1).

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