Fundamental Theorem of Calculus, Part 1

The Fundamental Theorem of Calculus, Part 1 is the most important theorem in calculus. It connects derivatives and integrals — the two operations that look completely different — and shows that they are essentially inverses of each other.

The Theorem

If $f$ is continuous on $[a, b]$ and we define the accumulation function

$F(x) = \int_{a}^{x} f(t)\,dt$

then $F$ is differentiable, and:

$F'(x) = f(x)$

In words: the rate of change of the accumulated area equals the integrand at the right endpoint.

This is why integrals "undo" derivatives: if you start with a function $f$, integrate it to get $F$, and then differentiate $F$, you get back the original $f$.

Step-by-Step Solution

Given: $F(x) = \displaystyle\int_{0}^{x} t^{2}\,dt$.

Find: $F'(x)$, and verify it equals $x^{2}$.

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Step 1 — Compute $F(x)$ directly using the antiderivative.

$F(x) = \int_{0}^{x} t^{2}\,dt = \dfrac{t^{3}}{3}\bigg|_{0}^{x} = \dfrac{x^{3}}{3} - 0 = \dfrac{x^{3}}{3}$

So $F(x) = x^{3}/3$ — a clean cubic.

Step 2 — Differentiate $F(x)$ using the power rule.

$F'(x) = \dfrac{d}{dx}\left[\dfrac{x^{3}}{3}\right] = \dfrac{1}{3} \cdot 3x^{2} = x^{2}$

So $F'(x) = x^{2}$.

Step 3 — Compare with the integrand.

The integrand is $f(t) = t^{2}$. At the upper limit, $f(x) = x^{2}$. So:

$F'(x) = x^{2} = f(x) \;\;\checkmark$

This is exactly what the FTC predicts.

Step 4 — Verify at specific points.

• At $x = 1$: $F(1) = 1/3$, $F'(1) = 1$, $f(1) = 1$ → match ✓
• At $x = 2$: $F(2) = 8/3 \approx 2.667$, $F'(2) = 4$, $f(2) = 4$ → match ✓
• At $x = 3$: $F(3) = 9$, $F'(3) = 9$, $f(3) = 9$ → match ✓

The slope of $F$ at any point exactly equals the value of the integrand at that point.

Step 5 — Why this works (intuition).

When you increase $x$ by a tiny amount $\Delta x$, the area $F$ increases by approximately a thin rectangle of height $f(x)$ and width $\Delta x$:

$F(x + \Delta x) - F(x) \approx f(x) \cdot \Delta x$

Dividing by $\Delta x$ and taking the limit gives:

$F'(x) = \lim_{\Delta x \to 0}\dfrac{F(x + \Delta x) - F(x)}{\Delta x} = f(x)$

That's the FTC in one limit.

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Answer: $\displaystyle F(x) = \int_{0}^{x} t^{2}\,dt = \dfrac{x^{3}}{3}$, and $F'(x) = x^{2}$. This equals the integrand $f(t) = t^{2}$ evaluated at the upper limit, exactly as the Fundamental Theorem of Calculus predicts.

$\boxed{\;\dfrac{d}{dx}\left[\int_{0}^{x} t^{2}\,dt\right] = x^{2}\;}$

Try It

• Slide the upper limit $x$ widget — watch the cyan integrand $t^{2}$ on the left, with the green shaded area showing $F(x)$.
• The right panel plots the accumulated area $F(x) = x^{3}/3$ growing with $x$, plus its derivative $F'(x) = x^{2}$ (pink).
• Notice that the slope of the green curve at each $x$ exactly matches the height of the cyan integrand at that $x$.