To determine
To evaluate:
∫01v2cos(v3)dv if it exists
Answer
13sin(1)
Explanation
1) Concept:
By using substitution rule for definite integrals and fundamental rule of calculus
2) Theorem:
Fundamental theorem of calculus:
If f is continuous on [a, b], then ∫abfxdx=Fb-F(a).
Substitution rule for definite integrals:
If g' is continuous on [a, b] and f is continuous on the range of u=g(x) then
∫abfgxg'xdx=∫g(a)g(b)f(u)du
3) Formula:
∫cosxdx=sinx+C
3) Given:
∫01v2cos(v3)dv
4) Calculation:
Consider, ∫01v2cos(v3)dv
Since the function is continuous on the given interval, the integral exists.
By applying substitution rule for definite integrals,
Let u=v3. Then du=3v2 dv, So v2dv=du3
When v=0, u=0 and v=1, u=1
Thus,
∫01v2cos(v3)dv=∫01cosu3du=13∫01cosudu
By using fundamental theorem of calculus and formula,
=13sinu01
Simplifying,
=13[sin1-sin0]
=13(sin1-0)
=13sin1
Conclusion:
Therefore, ∫01v2cos(v3)dv=13sin(1)