If
𝑓
is continuous and
∫
1
1024
𝑓
(
𝑡
)
𝑑
𝑡
=
83
,
evaluate the integral
∫
1
4
𝑡
4
𝑓
(
𝑡
5
)
𝑑
𝑡
.

Question

If 
𝑓
 is continuous and
∫
1
1024
𝑓
(
𝑡
)
𝑑
𝑡
=
83
,
evaluate the integral
∫
1
4
𝑡
4
𝑓
(
𝑡
5
)
𝑑
𝑡
.

BlackTom AI · Accepted Answer

Let u = t^5. Then du = 5t^4 dt, so t^4 dt = (1/5) du. This substitution transforms the given integral by changing both the integrand and the limits.

Relevant knowledge points:
- Change of variables (substitution) in definite integrals: replace t by a new variable u = g(t), convert dt using du = g'(t) dt, and change the integration limits accordingly.
- Continuity of f guarantees integrability and justifies the substitution and evaluation of the resulting definite integral.

Step-by-step reasoning and calculations:
1. Set u = t^5. Compute derivative: du/dt = 5 t^4, hence du = 5 t^4 dt and therefore t^4 dt = (1/5) du.
2. Convert limits: when t = 1, u = 1^5 = 1. When t = 4, u = 4^5 = 1024.
3. Substitute into the integral:
   ∫_{t=1}^{4} t^4 f(t^5) dt = ∫_{u=1}^{1024} f(u) * (1/5) du = (1/5) ∫_{1}^{1024} f(u) du.
4. Use the given value ∫_{1}^{1024} f(t) dt = 83. Thus the integral equals (1/5) * 83 = 83/5.

Why this is correct:
- The substitution directly matches the composition f(t^5) and the factor t^4 appearing in dt conversion, producing exactly (1/5) times the original integral over the transformed interval.
- Continuity of f ensures no pathological behavior; the integral remains well-defined under change of variables.

Why other answers would be incorrect:
- Multiplying instead of dividing by 5 (giving 5·83 = 415) would ignore the derivative factor du = 5 t^4 dt and thus misapply substitution.
- Any factor other than 1/5 (such as 1/25 or no factor) fails to account for du = 5 t^4 dt.

Conclusion: The value of the integral is 83/5, which matches the provided correct answer.

If 𝑓 is continuous and ∫ 1 1024 𝑓 ( 𝑡 ) 𝑑 𝑡 = 83 , evaluate the integral ∫ 1 4 𝑡 4 𝑓 ( 𝑡 5 ) 𝑑 𝑡 .Single choice

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