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:

∫cos⁡xdx=sin⁡x+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=∫01cos⁡u3du=13∫01cos⁡udu

By using fundamental theorem of calculus and formula,

=13sin⁡u01

Simplifying,

=13[sin⁡1-sin⁡0]

=13(sin⁡1-0)

=13sin⁡1

Conclusion:

Therefore, ∫01v2cos⁡(v3)dv=13sin⁡(1)