The Fundamental Theorem of Calculus for MATH 122
Exam Relevance for MATH 122
Tested both conceptually and computationally. Expect questions on evaluating definite integrals and FTC Part 2.
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Why is This Theorem So Important?
The Fundamental Theorem of Calculus (FTC) is arguably the most important theorem in all of calculus — and it has "Fundamental" right there in the name for a reason!
For centuries, mathematicians studied two seemingly unrelated problems:
- Tangent lines (slopes, rates of change) → led to derivatives
- Areas under curves (accumulation, totals) → led to integrals
The FTC reveals something shocking: these two problems are inverses of each other!
Differentiation and integration aren't just two random operations — they undo each other, like addition and subtraction, or multiplication and division. This connection is what makes calculus so powerful. Instead of approximating areas with tons of rectangles, you can find an anti-derivative and just plug in two numbers!
The Two Parts of the FTC
The Fundamental Theorem has two parts, and they say slightly different (but related) things.
Part 1: Derivatives of Integrals
FTC Part 1 tells us what happens when you differentiate an integral.
If $f$ is continuous on $[a, b]$, and we define:
$$g(x) = \int_a^x f(t)\,dt$$
then $g$ is differentiable and:
$$g'(x) = f(x)$$
In plain English: If you create a function by integrating from a fixed starting point $a$ up to a variable endpoint $x$, then the derivative of that function just gives you back the original integrand!
Why this matters: It proves that every continuous function has an anti-derivative. The integral itself constructs one for you.
Find $\frac{d}{dx}\left[\int_1^x t^3\,dt\right]$.
By FTC Part 1, we simply replace the dummy variable $t$ with $x$:
$$\frac{d}{dx}\left[\int_1^x t^3\,dt\right] = x^3$$
That's it! The derivative of the integral just returns the integrand with $x$ plugged in.
Find $\frac{d}{dx}\left[\int_0^{x^2} \cos(t)\,dt\right]$.
Here the upper limit is $x^2$, not just $x$. We need the chain rule!
Let $u = x^2$. By FTC Part 1:
$$\frac{d}{du}\left[\int_0^{u} \cos(t)\,dt\right] = \cos(u)$$
Now apply the chain rule (multiply by $\frac{du}{dx} = 2x$):
$$\frac{d}{dx}\left[\int_0^{x^2} \cos(t)\,dt\right] = \cos(x^2) \cdot 2x = 2x\cos(x^2)$$
Find $\frac{d}{dx}\left[\int_x^5 e^{t^2}\,dt\right]$.
The variable is in the lower limit, not the upper. We need to flip the integral first!
Using the property $\int_x^5 = -\int_5^x$:
$$\frac{d}{dx}\left[\int_x^5 e^{t^2}\,dt\right] = \frac{d}{dx}\left[-\int_5^x e^{t^2}\,dt\right]$$
$$= -\frac{d}{dx}\left[\int_5^x e^{t^2}\,dt\right]$$
Now apply FTC Part 1:
$$= -e^{x^2}$$
Find $\frac{d}{dx}\left[\int_{x}^{x^3} \sin(t)\,dt\right]$.
When both limits depend on $x$, split the integral at a constant (like 0):
$$\int_{x}^{x^3} \sin(t)\,dt = \int_{x}^{0} \sin(t)\,dt + \int_{0}^{x^3} \sin(t)\,dt$$
$$= -\int_{0}^{x} \sin(t)\,dt + \int_{0}^{x^3} \sin(t)\,dt$$
Now differentiate each part:
$$\frac{d}{dx}\left[-\int_{0}^{x} \sin(t)\,dt\right] = -\sin(x)$$
$$\frac{d}{dx}\left[\int_{0}^{x^3} \sin(t)\,dt\right] = \sin(x^3) \cdot 3x^2$$
Combining:
$$\frac{d}{dx}\left[\int_{x}^{x^3} \sin(t)\,dt\right] = -\sin(x) + 3x^2\sin(x^3)$$
Part 2: Evaluating Definite Integrals
FTC Part 2 tells us how to actually compute definite integrals.
If $f$ is continuous on $[a, b]$ and $F$ is any anti-derivative of $f$ (meaning $F'(x) = f(x)$), then:
$$\int_a^b f(x)\,dx = F(b) - F(a)$$
In plain English: To find the definite integral, find any anti-derivative, plug in the upper limit, plug in the lower limit, and subtract.
Why this matters: This is how you actually calculate integrals! Instead of adding up infinitely many rectangles, just find an anti-derivative and evaluate at two points.
Evaluate $\int_1^3 2x\,dx$.
Step 1: Find an anti-derivative
An anti-derivative of $2x$ is $x^2$.
Step 2: Apply FTC Part 2
$$\int_1^3 2x\,dx = x^2\Big|_1^3 = 3^2 - 1^2 = 9 - 1 = 8$$
Evaluate $\int_0^{\pi/4} \sec^2(x)\,dx$.
Step 1: Find an anti-derivative
An anti-derivative of $\sec^2(x)$ is $\tan(x)$.
Step 2: Apply FTC Part 2
$$\int_0^{\pi/4} \sec^2(x)\,dx = \tan(x)\Big|_0^{\pi/4} = \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1$$
Evaluate $\int_0^{\ln 2} e^x\,dx$.
Step 1: Find an anti-derivative
An anti-derivative of $e^x$ is $e^x$.
Step 2: Apply FTC Part 2
$$\int_0^{\ln 2} e^x\,dx = e^x\Big|_0^{\ln 2} = e^{\ln 2} - e^0 = 2 - 1 = 1$$
Conceptual Understanding
True or False: If $g(x) = \int_0^x f(t)\,dt$ and $f(t) > 0$ for all $t$ in $[0, 5]$, then $g(x)$ is increasing on $[0, 5]$.
True. By FTC Part 1, $g'(x) = f(x)$. Since $f(x) > 0$ on the interval, we have $g'(x) > 0$, which means $g$ is increasing. Geometrically, this makes sense: if you're always adding positive area, the accumulated total keeps growing.
True or False: If $g(x) = \int_0^x f(t)\,dt$ and $f$ is increasing, then $g$ is concave up.
True. By FTC Part 1, $g'(x) = f(x)$. Taking another derivative: $g''(x) = f'(x)$. If $f$ is increasing, then $f'(x) > 0$, so $g''(x) > 0$, which means $g$ is concave up.
True or False: $\frac{d}{dx}\left[\int_0^x \left(\int_0^t e^{s^2}\,ds\right)dt\right] = \int_0^x e^{s^2}\,ds$
True. Let's call the inner integral $h(t) = \int_0^t e^{s^2}\,ds$. Then we have $\frac{d}{dx}\left[\int_0^x h(t)\,dt\right]$. By FTC Part 1, this equals $h(x) = \int_0^x e^{s^2}\,ds$. The FTC "peels off" one layer of integration.
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting the chain rule in FTC Part 1
Wrong: $\frac{d}{dx}\left[\int_0^{x^2} t^3\,dt\right] = (x^2)^3 = x^6$
Why it's wrong: When the upper limit is a function of $x$ (not just $x$ itself), you must multiply by the derivative of that function.
Correct: $\frac{d}{dx}\left[\int_0^{x^2} t^3\,dt\right] = (x^2)^3 \cdot 2x = x^6 \cdot 2x = 2x^7$
❌ Mistake: Wrong sign when variable is in lower limit
Wrong: $\frac{d}{dx}\left[\int_x^3 t^2\,dt\right] = x^2$
Why it's wrong: When the variable is in the lower limit, you need to flip the integral first, which introduces a negative sign.
Correct: $\frac{d}{dx}\left[\int_x^3 t^2\,dt\right] = \frac{d}{dx}\left[-\int_3^x t^2\,dt\right] = -x^2$
❌ Mistake: Confusing the dummy variable with the limit
Wrong: $\frac{d}{dx}\left[\int_1^x t^2\,dt\right] = t^2$
Why it's wrong: The $t$ is a "dummy variable" — it disappears after integration. The answer should be in terms of $x$.
Correct: $\frac{d}{dx}\left[\int_1^x t^2\,dt\right] = x^2$
Fundamental Theorem of Calculus (Part 1)
The derivative of an integral (with variable upper limit) gives back the original function. Differentiation undoes integration!
Variables:
- $a$:
- fixed lower limit of integration
- $x$:
- variable upper limit
- $f(t)$:
- continuous function being integrated
- $t$:
- dummy variable (gets replaced by x after differentiation)
FTC Part 1 with Chain Rule
When the upper limit is a function of x (not just x), multiply by the derivative of that function.
Variables:
- $g(x)$:
- function in the upper limit
- $g'(x)$:
- derivative of the upper limit function
- $f(g(x))$:
- original integrand with g(x) plugged in
Fundamental Theorem of Calculus (Part 2)
To evaluate a definite integral, find an anti-derivative F, plug in the upper and lower limits, and subtract. Integration undoes differentiation!
Variables:
- $f(x)$:
- the function being integrated
- $F(x)$:
- any anti-derivative of f(x), where F'(x) = f(x)
- $a$:
- lower limit of integration
- $b$:
- upper limit of integration
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