Calculate the integral of (x^3-4x^2+1) from 1 to 2 using the fundamental theorem of calculus.



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Posted on (Answer #1)

Fundamental theorem of calculus says:

Let `f` be continous real-valued function defined on segment `[a,b]` and let `F` be defined as

`F=int_a^x f(t)dt.`

Then `F` is continous and differentiable on `(a,b)` and for all `x in (a,b)`


For calculating definite integral we usually use corollary of this theorem, also known as Newton-Leibniz formula

Same assumptions as in previous theorem and


This is sometimes called second fundamental theorem of calculus.

Now to calculate your integral:

`int_1^2(x^3-4x^2+1)dx=` by linearity of integral



Also see the link below.

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