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Day 19

Fundamental Theorem Of Calculus

The Fundamental Theorem of Calculus, Part 1The Fundamental Theorem of Calculus, Part 2U-Substitution: Reversing the Chain Rule

The Fundamental Theorem of Calculus, Part 1

Evaluating a definite integral with an antiderivative. If F′=fF' = f, then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) — this turns limit-of-Riemann-sum calculations into simple algebra.

Core Theorem
If ff is continuous on [a,b][a, b] and FF is any antiderivative of ff (that is, F′(x)=f(x)F'(x) = f(x)), then
∫abf(x) dx=F(b)−F(a)=[F(x)]ab\int_a^b f(x) \, dx = F(b) - F(a) = \Big[F(x)\Big]_a^b
Step-by-Step SOP
  1. 1

    Find F(x)

    Determine an antiderivative of f(x)f(x).
  2. 2

    Evaluate at the bounds

    Compute F(b)F(b) and F(a)F(a).
  3. 3

    Subtract

    The value of the integral is F(b)−F(a)F(b) - F(a).

Practice Exercises

Example 01Easy
Evaluate ∫13x2 dx\int_1^3 x^2 \, dx.
NEED A HINT?
An antiderivative of x2x^2 is x33\frac{x^3}{3}.
SHOW DETAILED EXPLANATION

Find an antiderivative

F(x)=x33F(x) = \frac{x^3}{3}, since F′(x)=x2F'(x) = x^2.

Evaluate at the endpoints

F(3)=273=9F(3) = \frac{27}{3} = 9 and F(1)=13F(1) = \frac{1}{3}.

Subtract

∫13x2 dx=9−13=263\int_1^3 x^2 \, dx = 9 - \frac{1}{3} = \frac{26}{3}.
Example 02Medium
Evaluate ∫0πsin⁡(x) dx\int_0^{\pi} \sin(x) \, dx.
NEED A HINT?
The antiderivative of sin⁡(x)\sin(x) is −cos⁡(x)-\cos(x).
SHOW DETAILED EXPLANATION

Antiderivative

F(x)=−cos⁡(x)F(x) = -\cos(x).

Evaluate

F(π)=−cos⁡(π)=1F(\pi) = -\cos(\pi) = 1 and F(0)=−cos⁡(0)=−1F(0) = -\cos(0) = -1.

Subtract

1−(−1)=21 - (-1) = 2.
Example 03Medium
Evaluate ∫01(ex+2x) dx\int_0^1 (e^x + 2x) \, dx.
NEED A HINT?
Integrate exe^x and 2x2x separately using the sum rule.
SHOW DETAILED EXPLANATION

Split the integral

∫01ex dx+∫012x dx\int_0^1 e^x \, dx + \int_0^1 2x \, dx.

Antiderivatives

F(x)=ex+x2F(x) = e^x + x^2.

Evaluate

(e1+12)−(e0+02)=(e+1)−1=e(e^1 + 1^2) - (e^0 + 0^2) = (e + 1) - 1 = e.

The Fundamental Theorem of Calculus, Part 2

Differentiating a function defined by an integral. If g(x)=∫axf(t) dtg(x) = \int_a^x f(t)\,dt, then g′(x)=f(x)g'(x) = f(x) — differentiation undoes integration. With a function in the upper limit, multiply by its derivative (chain rule).

Core Theorem
If ff is continuous and g(x)=∫axf(t) dtg(x) = \int_a^x f(t) \, dt, then g′(x)=f(x)g'(x) = f(x). More generally, with a function u(x)u(x) in the upper limit:
ddx∫au(x)f(t) dt=f(u(x))⋅u′(x)\frac{d}{dx} \int_a^{u(x)} f(t) \, dt = f\big(u(x)\big) \cdot u'(x)

Practice Exercises

Example 01Easy
Find ddx∫0xsin⁡(t) dt\frac{d}{dx} \int_0^x \sin(t) \, dt.
NEED A HINT?
Apply FTC Part 2 directly — the upper limit is just xx.
SHOW DETAILED EXPLANATION

Recognize the form

This is ∫axf(t) dt\int_a^x f(t)\,dt with f(t)=sin⁡(t)f(t) = \sin(t).

Apply the theorem

ddx∫0xsin⁡(t) dt=sin⁡(x)\frac{d}{dx} \int_0^x \sin(t) \, dt = \sin(x).
Example 02Medium
Find ddx∫1x2et dt\frac{d}{dx} \int_1^{x^2} e^t \, dt.
NEED A HINT?
The upper limit is u(x)=x2u(x) = x^2, so use the chain-rule form.
SHOW DETAILED EXPLANATION

Identify f and u

f(t)=etf(t) = e^t, u(x)=x2u(x) = x^2, so u′(x)=2xu'(x) = 2x.

Apply the rule

f(u(x))⋅u′(x)=ex2⋅2x=2x ex2f\big(u(x)\big)\cdot u'(x) = e^{x^2}\cdot 2x = 2x\,e^{x^2}.
Example 03Medium
If g(x)=∫2x(3t2+1) dtg(x) = \int_2^{x} (3t^2 + 1)\,dt, find g′(x)g'(x) and g(2)g(2).
NEED A HINT?
FTC Part 2 gives g′g' instantly. For g(2)g(2), look at the interval of integration.
SHOW DETAILED EXPLANATION

Differentiate

g′(x)=3x2+1g'(x) = 3x^2 + 1.

Evaluate g(2)

g(2)=∫22(3t2+1) dt=0g(2) = \int_2^2 (3t^2 + 1)\,dt = 0 (zero-width interval).
Common Pitfalls
  • ⚠
    Forgetting the chain-rule factorWhen the upper limit is x2x^2 (not just xx), you must multiply by 2x2x. Leaving it out is the most common mistake.
  • ⚠
    Variable in the lower limitIf xx is the lower limit, ddx∫xbf(t) dt=−f(x)\frac{d}{dx}\int_x^b f(t)\,dt = -f(x) — flip the sign first using the reversal property.

U-Substitution: Reversing the Chain Rule

The integration counterpart of the chain rule. When the integrand contains a function and (a multiple of) its derivative, substitute uu for the inner function to collapse the integral into a basic form.

Core Theorem
If u=g(x)u = g(x) so that du=g′(x) dxdu = g'(x)\,dx, then
∫f(g(x)) g′(x) dx=∫f(u) du\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du
For a definite integral, either convert the limits to uu-values or convert back to xx before evaluating.
Step-by-Step SOP
  1. 1

    Pick u

    Choose uu as the inner function — often what sits inside a power, root, exponential, or denominator.
  2. 2

    Compute du

    Differentiate: du=g′(x) dxdu = g'(x)\,dx. Solve for whatever piece the integrand needs.
  3. 3

    Substitute

    Rewrite the whole integral in uu and dudu — no xx should remain.
  4. 4

    Integrate

    Evaluate the simpler integral in uu.
  5. 5

    Convert back

    Replace uu with g(x)g(x) for an indefinite integral, or change the limits for a definite one.

Practice Exercises

Example 01Easy
Evaluate ∫2x (x2+1)5 dx\int 2x\,(x^2 + 1)^5 \, dx.
NEED A HINT?
Let u=x2+1u = x^2 + 1. Then du=2x dxdu = 2x\,dx, which is already present.
SHOW DETAILED EXPLANATION

Choose u

u=x2+1  ⟹  du=2x dxu = x^2 + 1 \implies du = 2x\,dx.

Rewrite in u

∫u5 du\int u^5 \, du.

Integrate

u66+C\frac{u^6}{6} + C.

Substitute back

(x2+1)66+C\frac{(x^2 + 1)^6}{6} + C.
Example 02Medium
Evaluate ∫xx2+4 dx\int x\sqrt{x^2 + 4} \, dx.
NEED A HINT?
Let u=x2+4u = x^2 + 4, so du=2x dxdu = 2x\,dx and x dx=12 dux\,dx = \tfrac{1}{2}\,du.
SHOW DETAILED EXPLANATION

Choose u and solve for the piece you need

u=x2+4  ⟹  du=2x dx  ⟹  x dx=12 duu = x^2 + 4 \implies du = 2x\,dx \implies x\,dx = \tfrac{1}{2}\,du.

Rewrite in u

∫u⋅12 du=12∫u1/2 du\int \sqrt{u}\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}\int u^{1/2}\,du.

Integrate

12⋅u3/23/2=u3/23\tfrac{1}{2}\cdot\frac{u^{3/2}}{3/2} = \frac{u^{3/2}}{3}.

Substitute back

(x2+4)3/23+C\frac{(x^2 + 4)^{3/2}}{3} + C.
Example 03Medium
Evaluate ∫01xx2+1 dx\int_0^1 \frac{x}{x^2 + 1} \, dx.
NEED A HINT?
Let u=x2+1u = x^2 + 1 and change the limits: x=0⇒u=1x=0 \Rightarrow u=1, x=1⇒u=2x=1 \Rightarrow u=2.
SHOW DETAILED EXPLANATION

Substitute and change the limits

u=x2+1u = x^2 + 1, du=2x dxdu = 2x\,dx; the limits become u:1→2u: 1 \to 2.

Rewrite in u

∫121u⋅12 du=12∫121u du\int_1^2 \frac{1}{u}\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}\int_1^2 \frac{1}{u}\,du.

Integrate and evaluate

12[ln⁡∣u∣]12=12(ln⁡2−ln⁡1)=12ln⁡2\tfrac{1}{2}\Big[\ln|u|\Big]_1^2 = \tfrac{1}{2}(\ln 2 - \ln 1) = \tfrac{1}{2}\ln 2.
Example 04Hard
Evaluate ∫0π/2cos⁡(x) esin⁡(x) dx\int_0^{\pi/2} \cos(x)\,e^{\sin(x)} \, dx.
NEED A HINT?
Let u=sin⁡(x)u = \sin(x), so du=cos⁡(x) dxdu = \cos(x)\,dx. Change the limits.
SHOW DETAILED EXPLANATION

Substitute

u=sin⁡(x)u = \sin(x); the limits x:0→π/2x: 0 \to \pi/2 become u:0→1u: 0 \to 1.

Rewrite and integrate

∫01eu du=[eu]01=e−1\int_0^1 e^u \, du = \Big[e^u\Big]_0^1 = e - 1.
Common Pitfalls
  • ⚠
    Not changing the limits on a definite integralIf you switch to uu but keep the original xx-limits, the answer is wrong. Either convert the limits to uu-values, or convert back to xx before evaluating.
  • ⚠
    Leftover x termsAfter substituting, every xx must be gone. If an xx remains, the substitution is wrong or you also need to solve u=g(x)u = g(x) for xx and substitute that.
Gary Chang

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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