Differentiate an Accumulation Function
If a continuous function is accumulated from a fixed lower bound to variable upper input x, the derivative of that accumulation equals the integrand evaluated at x.
About 25 minutes, plus any extended practice
Supplied materials
- Paper or a browser scratch area; a calculator is optional unless rounding is requested.
- Supplied teaching case: Let F(x)=integral from 0 to x of (t²+1) dt. Find F′(2). Candidate answers: (A) 10 (B) 4 (C) 5 (D) 6.
- Independent check inputs: Let F(x)=integral from 0 to x of (t²+1) dt. Find F′(3). Candidate answers: (A) 10 (B) 11 (C) 20 (D) 9.
Learn the idea
Integration connects signed accumulation with limiting sums. Rectangle models approximate that accumulation, antiderivatives evaluate suitable definite integrals, and the Fundamental Theorem links an accumulation function’s derivative to its continuous integrand. If a continuous function is accumulated from a fixed lower bound to variable upper input x, the derivative of that accumulation equals the integrand evaluated at x. This is a Fundamental Theorem relationship.
Here is the method on the supplied teaching case. The integrand is continuous, so the Fundamental Theorem gives F′(x)=x²+1. At 2, this equals 5. No integration constant or extra power is needed.
Continuity is satisfied by the supplied polynomial integrand. Do not confuse differentiating an accumulation with integrating again; a varying lower bound or a composed upper bound would require additional rules outside this simple case.
- Write or sketch the supplied teaching case, labeling its quantities and restrictions.
- Apply the method shown in the worked solution, keeping each intermediate equality or justified inference.
- Solve the independent check with its changed inputs before revealing feedback.
- Complete the three practice cases; explain which mathematical rule rules out one distractor in each case.
See the method in action
Teaching case: Let F(x)=integral from 0 to x of (t²+1) dt. Find F′(2). Candidate answers: (A) 10 (B) 4 (C) 5 (D) 6. Solution: The integrand is continuous, so the Fundamental Theorem gives F′(x)=x²+1. At 2, this equals 5. No integration constant or extra power is needed. The final answer is 5. Continuity is satisfied by the supplied polynomial integrand. Do not confuse differentiating an accumulation with integrating again; a varying lower bound or a composed upper bound would require additional rules outside this simple case.
Check your reasoning
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Helpful tips
- If a continuous function is accumulated from a fixed lower bound to variable upper input x, the derivative of that accumulation equals the integrand evaluated at x.
- Continuity is satisfied by the supplied polynomial integrand.
See the answer and explanation
10
The integrand is continuous, so the Fundamental Theorem gives F′(x)=x²+1. At 3, this equals 10. No integration constant or extra power is needed.
Take it further
Changed-input task: Let F(x)=integral from 0 to x of (t²+1) dt. Find F′(9). Candidate answers: (A) 164 (B) 81 (C) 82 (D) 83. Show the mathematical reason, not only the selected answer. Expected result: 82. The integrand is continuous, so the Fundamental Theorem gives F′(x)=x²+1. At 9, this equals 82. No integration constant or extra power is needed.
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