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