If
𝑓
(
−
1
)
=
−
2
,
𝑓
′
is continuous, and
∫
−
1
13
𝑓
′
(
𝑡
)
𝑑
𝑡
=
−
5
, what is the value of
𝑓
(
13
)
?

Question

If 
𝑓
(
−
1
)
=
−
2
, 
𝑓
′
 is continuous, and 
∫
−
1
13
𝑓
′
(
𝑡
)
𝑑
𝑡
=
−
5
, what is the value of 
𝑓
(
13
)
?

BlackTom AI · Accepted Answer

The Fundamental Theorem of Calculus states that the integral of a derivative over an interval equals the net change of the original function over that interval: f(b) − f(a) = ∫_a^b f′(t) dt. Because f′ is continuous, it is integrable and the theorem applies directly.

1) Relevant knowledge points and concepts
- Fundamental Theorem of Calculus (net change form): f(13) − f(−1) = ∫_{−1}^{13} f′(t) dt.
- Continuity of f′ ensures the integral exists and the theorem is valid.
- Given value f(−1) = −2 and ∫_{−1}^{13} f′(t) dt = −5.

2) Step-by-step reasoning
Apply the net-change formula:
f(13) − f(−1) = ∫_{−1}^{13} f′(t) dt.

3) Calculations with intermediate steps
Substitute the given integral:
f(13) − f(−1) = −5.
Now substitute f(−1) = −2:
f(13) − (−2) = −5
which is f(13) + 2 = −5.
Solve for f(13):
f(13) = −5 − 2 = −7.

4) Why the correct answer is correct
The result follows directly from the Fundamental Theorem: the definite integral of f′ over [−1,13] equals the change in f over that interval. Adding that change (−5) to the known starting value f(−1) = −2 yields f(13) = −7.

5) Why other answers would be incorrect
Common incorrect mistakes would be (a) adding instead of adding the negative (−2 + 5 = 3) — incorrect because the integral is −5, not +5; (b) forgetting to add f(−1) and taking f(13) = −5 alone — incorrect because the integral gives change, not absolute value; (c) sign errors in algebra when isolating f(13). Each of these missteps fails to apply the net-change formula or mishandles signs.

6) Conclusion
Therefore f(13) = −7, which matches the provided correct answer (−7).

If 𝑓 ( − 1 ) = − 2 , 𝑓 ′ is continuous, and ∫ − 1 13 𝑓 ′ ( 𝑡 ) 𝑑 𝑡 = − 5 , what is the value of 𝑓 ( 13 ) ?单项选择题

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