Understand the first fundamental theorem through small added intervals and variable bounds.
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If a tank has accumulated water since a fixed starting time, extending the observation by a short duration adds approximately current flow times that duration. Dividing the added volume by the duration recovers an average flow over the short interval. Predict what happens as the interval shrinks when the flow is continuous.
Predict
Let A(x)=∫₁ˣ(t²+2)dt. Predict whether A is increasing even when x is smaller than 1 and A is negative. A function’s value and rate can have different signs. The dummy variable t names points being accumulated; x determines where accumulation stops.
Experiment
Use the integrand values to determine rates of accumulation without first finding a formula for A.
Endpoint x
Integrand t²+2 evaluated at x
A′(x)
0
2
2
1
3
3
3
11
11
Draw a positive integrand and shade the interval from 1 to x. Then move x a small distance right. The newly added thin strip explains why A′ is positive everywhere, while the reversed orientation for x below 1 explains negative accumulated values there.
Understand
The first fundamental theorem of calculus, in its continuous-integrand form, states that A(x)=∫ₐˣf(t)dt has derivative A′(x)=f(x) at interior points where f is continuous. The lower endpoint is fixed. This theorem constructs an antiderivative through accumulation, even when no elementary symbolic formula for that antiderivative is available.
Subtract A(x) from A(x+h). Integral additivity leaves only ∫ₓ^(x+h)f(t)dt. Dividing by h gives an average over a short interval. The difference between this quotient and f(x) is bounded by the largest nearby value of |f(t)−f(x)|. Continuity makes that bound tend to zero. For negative h, both the orientation and divisor reverse, giving the same averaging argument. This explains the derivative rather than merely naming a rule.
A moving endpoint needs a chain factor. For B(x)=∫₀^(x²)(1+t)dt, first view it as A(x²), then differentiate to obtain (1+x²)·2x. At x=2 the rate is 20, not 5. The integrand value describes rate per endpoint unit; the chain factor describes endpoint change per x unit.
If both bounds move, split through any fixed reference. For C(x)=∫ₓ^(x²)e^t dt, write C=F(x²)−F(x) with F′=e^x. Therefore C′=2x e^(x²)−e^x. The lower-endpoint term is negative because advancing the start removes a contribution. This identity also handles reversed endpoint order consistently.
Continuity matters. Let f(t)=−1 for negative t and 1 for nonnegative t. Then ∫₀ˣf(t)dt=|x|, whose left and right slopes at zero disagree. The value assigned to f at the single jump does not repair the accumulated function’s corner. Do not apply the continuous-integrand theorem at that point as though its hypothesis held.
Apply
Practice interpreting the sign of A′ from the integrand graph. A negative rate means the accumulated quantity decreases; it does not necessarily mean the accumulated quantity is negative. In variable-bound problems, name the endpoint functions and differentiate each one explicitly. In the proof task, explain the role of continuity in controlling a shrinking average.
Continue
Next use antiderivatives to evaluate definite integrals. The two fundamental-theorem directions are related but answer different questions: one differentiates accumulation, while the other computes it from endpoint values.
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Practice and reasoning
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Let A(x)=∫₁ˣ (t²+2) dt. Find A′(3).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Let B(x)=∫₀^(x²) (1+t) dt. Find B′(2).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
If A(x)=∫₀ˣ f(t) dt and continuous f is negative throughout (1,2), then A on that interval is:
Explain why A′(x)=f(x) for A(x)=∫ₐˣ f(t)dt when f is continuous near x, using the increment quotient.
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Differentiate C(x)=∫ₓ^(x²) e^t dt.
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Let f(t)=−1 for t<0 and f(t)=1 for t≥0. With A(x)=∫₀ˣ f(t)dt, explain why A is not differentiable at zero.
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Reflect and continue
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