To determine

To evaluate:

∫03x2-4 dx if it exists

Answer

∫03x2-4 dx=233

Explanation

1) Concept:

By using the Fundamental Theorem of calculus and rules of integration, evaluate the given integral.

2) Theorem:

Fundamental theorem of calculus:

If f is continuous on [a, b], then ∫abfxdx=Fb-F(a).

3) Formula:

∫xndx=xn+1n+1+C

4) Given:

∫03x2-4 dx

5) Calculation:

Consider, ∫03x2-4 dx

x2-4= -(x2-4),       0≤x<2(x2-4),       2<x≤3

x2-4= (4-x2),       0≤x<2(x2-4),       2<x≤3

Since the function continuous in the given interval, the integral exists. Now

∫03x2-4 dx= ∫02(4-x2) dx+∫23(x2-4) dx

By applying fundamental theorem of calculus and power rule,

=4x-x3302+ x33-4x23

=4(2)-233-4(0)-033+333-4(3)-233-4(2)

=8-83- 0+273-12-83-8

=8-83+9-12-83+8

=13-163

=39-163

=233

Conclusion:

Therefore,

∫03x2-4 dx=233