18

Day 18

The Definite Integral And Ftc


Properties of the Definite Integral: Linearity, Additivity, and Order

Just as derivatives have a sum rule and a constant-multiple rule, integrals have algebraic properties that let you break a hard integral into easy pieces or combine known results — especially when the integrand is only given by a graph or a few known integral values.

Core Theorem
For integrable functions f,gf, g and constants c,a,bc, a, b:
1. Sum rule:
∫ab[f(x)+g(x)] dx=∫abf(x) dx+∫abg(x) dx\int_a^b [f(x) + g(x)] \, dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx
2. Constant multiple:
∫abc f(x) dx=c∫abf(x) dx\int_a^b c\,f(x)\,dx = c \int_a^b f(x)\,dx
3. Additivity over intervals:
∫acf(x) dx+∫cbf(x) dx=∫abf(x) dx\int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \int_a^b f(x)\,dx
4. Reversing the limits:
∫abf(x) dx=−∫baf(x) dx\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx
5. Zero width:
∫aaf(x) dx=0\int_a^a f(x)\,dx = 0

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy
Given ∫02f(x) dx=5\int_0^2 f(x)\,dx = 5 and ∫02g(x) dx=3\int_0^2 g(x)\,dx = 3, find ∫02[2f(x)−g(x)] dx\int_0^2 [2f(x) - g(x)]\,dx.
How to think about it
You often can't compute these integrals — you combine known values. Integration is linear, so split over sums and pull constants out; and it's additive, so adjacent intervals join up.

Key idea: Use the constant-multiple rule and the sum rule.

Full solution

1. Split using linearity

∫02[2f(x)−g(x)] dx=2∫02f(x) dx−∫02g(x) dx\int_0^2 [2f(x) - g(x)]\,dx = 2\int_0^2 f(x)\,dx - \int_0^2 g(x)\,dx.

2. Substitute the given values

=2(5)−3=7= 2(5) - 3 = 7.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium
Given ∫03h(x) dx=8\int_0^3 h(x)\,dx = 8 and ∫53h(x) dx=2\int_5^3 h(x)\,dx = 2, find ∫05h(x) dx\int_0^5 h(x)\,dx.
Need a hint?
∫53h(x) dx\int_5^3 h(x)\,dx has its limits reversed relative to ∫35h(x) dx\int_3^5 h(x)\,dx.
Show solution

1. Fix the reversed limits

∫53h(x) dx=−∫35h(x) dx=2  ⟹  ∫35h(x) dx=−2\int_5^3 h(x)\,dx = -\int_3^5 h(x)\,dx = 2 \implies \int_3^5 h(x)\,dx = -2.

2. Combine with additivity

∫05h(x) dx=∫03h(x) dx+∫35h(x) dx=8+(−2)=6\int_0^5 h(x)\,dx = \int_0^3 h(x)\,dx + \int_3^5 h(x)\,dx = 8 + (-2) = 6.
Practice 02Medium
The graph of ff on [0,6][0, 6] encloses a triangle of area 4 above the xx-axis on [0,3][0, 3] and a region of area 5 below the xx-axis on [3,6][3, 6]. Find ∫06f(x) dx\int_0^6 f(x)\,dx and ∫06∣f(x)∣ dx\int_0^6 |f(x)|\,dx.
Need a hint?
The plain integral uses signed area; the absolute-value integral uses total area.
Show solution

1. Signed area

∫06f(x) dx=(+4)+(−5)=−1\int_0^6 f(x)\,dx = (+4) + (-5) = -1.

2. Total area

∫06∣f(x)∣ dx=4+5=9\int_0^6 |f(x)|\,dx = 4 + 5 = 9.
Common Pitfalls
  • ⚠
    Assuming $\int (fg) = (\int f)(\int g)$There is no product rule for integrals. Linearity applies only to sums and constant multiples.

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

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy
Evaluate ∫13x2 dx\int_1^3 x^2 \, dx.
How to think about it
The whole trick is: find any antiderivative FF, then the definite integral is just F(b)−F(a)F(b) - F(a). No Riemann sums, no limits — the hard work is finding FF.

Key idea: An antiderivative of x2x^2 is x33\frac{x^3}{3}.

Full solution

1. Find an antiderivative

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

2. Evaluate at the endpoints

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

3. Subtract

∫13x2 dx=9−13=263\int_1^3 x^2 \, dx = 9 - \frac{1}{3} = \frac{26}{3}.
Worked Example 02Medium
Evaluate ∫0πsin⁡(x) dx\int_0^{\pi} \sin(x) \, dx.
How to think about it

Key idea: The antiderivative of sin⁡(x)\sin(x) is −cos⁡(x)-\cos(x).

Full solution

1. Antiderivative

F(x)=−cos⁡(x)F(x) = -\cos(x).

2. Evaluate

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

3. Subtract

1−(−1)=21 - (-1) = 2.
Step-by-Step SOP
  1. 1

    Find F(x)

    Determine an antiderivative of f(x)f(x) (rewrite the integrand first if needed).
  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 — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium
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 solution

1. Split the integral

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

2. Antiderivatives

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

3. Evaluate

(e1+12)−(e0+02)=(e+1)−1=e(e^1 + 1^2) - (e^0 + 0^2) = (e + 1) - 1 = e.
Practice 02Medium
Evaluate ∫123x2−1x dx\int_1^2 \dfrac{3x^2 - 1}{x}\,dx.
Need a hint?
Divide through by xx first: 3x−1x3x - \frac{1}{x}.
Show solution

1. Rewrite the integrand

3x2−1x=3x−1x\dfrac{3x^2 - 1}{x} = 3x - \dfrac{1}{x}.

2. Antiderivative

F(x)=3x22−ln⁡∣x∣F(x) = \dfrac{3x^2}{2} - \ln|x|.

3. Evaluate

(6−ln⁡2)−(32−0)=92−ln⁡2\left(6 - \ln 2\right) - \left(\dfrac{3}{2} - 0\right) = \dfrac{9}{2} - \ln 2.
Common Pitfalls
  • ⚠
    Trying to Evaluate Without an AntiderivativeFTC Part 1 needs FF with F′=fF' = f. If you cannot find one by hand, the AP context expects a calculator (Part A) or a geometry/accumulation argument.

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)

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy
Find ddx∫0xsin⁡(t) dt\frac{d}{dx} \int_0^x \sin(t) \, dt.
How to think about it
Differentiating an integral undoes it: ddx∫axf(t) dt=f(x)\frac{d}{dx}\int_a^x f(t)\,dt = f(x). If the top limit is a function u(x)u(x) rather than plain xx, tack on ⋅ u′(x)\cdot\, u'(x) (chain rule).

Key idea: Apply FTC Part 2 directly — the upper limit is just xx.

Full solution

1. Recognize the form

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

2. Apply the theorem

ddx∫0xsin⁡(t) dt=sin⁡(x)\frac{d}{dx} \int_0^x \sin(t) \, dt = \sin(x).
Worked Example 02Medium
Find ddx∫1x2et dt\frac{d}{dx} \int_1^{x^2} e^t \, dt.
How to think about it

Key idea: The upper limit is u(x)=x2u(x) = x^2, so use the chain-rule form.

Full solution

1. Identify f and u

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

2. 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}.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium
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 solution

1. Differentiate

g′(x)=3x2+1g'(x) = 3x^2 + 1.

2. Evaluate g(2)

g(2)=∫22(3t2+1) dt=0g(2) = \int_2^2 (3t^2 + 1)\,dt = 0 (zero-width interval).
Practice 02Hard
Find ddx∫x4cos⁡(t2) dt\frac{d}{dx} \int_x^{4} \cos(t^2)\,dt.
Need a hint?
The variable is in the LOWER limit. Reverse the limits (introducing a minus sign) so xx is on top.
Show solution

1. Reverse the limits

∫x4cos⁡(t2) dt=−∫4xcos⁡(t2) dt\int_x^4 \cos(t^2)\,dt = -\int_4^x \cos(t^2)\,dt.

2. Differentiate

ddx[−∫4xcos⁡(t2) dt]=−cos⁡(x2)\frac{d}{dx}\left[-\int_4^x \cos(t^2)\,dt\right] = -\cos(x^2).
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.
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.