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

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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)`

`F'(x)=f(x)`

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

Same assumptions as in previous theorem and

`int_a^bf(x)dx=F(b)-F(a)`

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

`int_1^2x^3dx-4int_1^2x^2dx+int_1^2dx=x^4/4|_1^2-4/3x^3|_1^2+x|_1^2=`

`4-1/4-32/3+4/3+2-1=-55/12`

Also see the link below.

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