Single-Variable-Calculus · Unit 15 · Video 1 · Interactive Practice

The Fundamental Theorem of Calculus: Area in One Line

IKey Formulas

Formula	Name	What you need
$\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$	Fundamental theorem (FTC 1)	$f$ continuous and $F'(x) = f(x)$
$F(x)\,\Big\|_a^b = F(b) - F(a)$	Evaluation bar	Top value minus bottom value
$\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$	Power rule for antiderivatives	Any power with $n \ne -1$
$\big(F(b) + c\big) - \big(F(a) + c\big) = F(b) - F(a)$	Any antiderivative works	The constant cancels

Key Insight: An area problem becomes an antiderivative problem: $\displaystyle\int_0^b x^2\,dx = \frac{x^3}{3}\,\Big|_0^b = \frac{b^3}{3}$ , in one line instead of a limit of Riemann sums.

IIArea Under $f$ Equals the Rise of $F$

The area under $f(x) = x^2$ from $a$ to $b$ equals $F(b) - F(a)$ for every antiderivative $F(x) = \tfrac{x^3}{3} + c$ .

Lower limit a 0.4 Upper limit b 1.1 Constant c 0

IIIOne Hump of Sine

How large is the area under one hump of $\sin x$ , from $0$ to $\pi$ ?

Your guess for the area 1.0

IVThe Power Rule on $[0, 1]$

For every $n \ge 0$ , $\int_0^1 x^n\,dx = \frac{1}{n+1}$ : the later the curve climbs, the smaller the area.

Exponent n 100

💡 With Riemann sums, the case $n = 100$ would have meant adding up $1^{100} + 2^{100} + \dots + k^{100}$ for $k$ rectangles and taking a limit.

VQuiz Questions

Problem 1 · Top Minus Bottom

Find $\displaystyle\int_1^3 x^2\,dx$ .

✅ Correct!

\tfrac{x^3}{3}\big|_1^3 = 9 - \tfrac{1}{3} = \tfrac{26}{3}

❌ Not quite.

9

is only the top value

F(3) = \tfrac{27}{3}

. The lower limit is

1

, not

0

, so

F(1) = \tfrac{1}{3}

must be subtracted.

❌ Not quite.

3^2 - 1^2 = 8

evaluates the integrand

f

itself. The theorem evaluates an antiderivative

F

with

F' = f

❌ Check the operation. The bar means top value minus bottom value, not plus.

❌ Not quite. Use

F(x) = \tfrac{x^3}{3}

and compute

F(3) - F(1)

Show solution

An antiderivative of $x^2$ is $F(x) = \tfrac{x^3}{3}$ , since $F'(x) = \tfrac{3x^2}{3} = x^2$ . By the fundamental theorem:

\int_1^3 x^2\,dx = \frac{x^3}{3}\,\Big|_1^3 = \frac{3^3}{3} - \frac{1^3}{3} = 9 - \frac{1}{3} = \frac{26}{3}.

Problem 2 · Watch the Signs

Find $\displaystyle\int_0^{\pi/3} \sin x\,dx$ .

✅ Correct!

-\cos\tfrac{\pi}{3} - (-\cos 0) = -\tfrac{1}{2} + 1 = \tfrac{1}{2}

. A positive function over

[0, \tfrac{\pi}{3}]

must give a positive area.

❌ Sign error in the antiderivative. The derivative of

\cos x

-\sin x

, so the antiderivative of

\sin x

-\cos x

, not

\cos x

❌ Watch the bottom value. The bottom value is

-\cos 0 = -1

, and subtracting it gives

-(-1) = +1

. Dropping the parentheses turned it into

-1

❌ Not quite.

\sin\tfrac{\pi}{3} - \sin 0

evaluates the integrand. Evaluate the antiderivative

-\cos x

instead.

❌ Not quite. Compute

-\cos x\,\big|_0^{\pi/3}

slowly, one sign at a time.

Show solution

Since $(-\cos x)' = \sin x$ , take $F(x) = -\cos x$ :

\int_0^{\pi/3} \sin x\,dx = -\cos x\,\Big|_0^{\pi/3} = -\cos\frac{\pi}{3} - (-\cos 0) = -\frac{1}{2} - (-1) = \frac{1}{2}.

Check: the region sits under the hump, starting at $0$ , so the answer must be positive and less than the full hump area $2$ .

Problem 3 · Power Rule, Nonzero Lower Limit

Given: $\displaystyle\int_1^2 x^4\,dx$ — find an antiderivative of $x^4$ and the value of the integral.

Which function is an antiderivative of $x^4$ ?

What is the value of the integral?

✅ Correct!

\tfrac{x^5}{5}\big|_1^2 = \tfrac{32}{5} - \tfrac{1}{5} = \tfrac{31}{5}

❌ Check by differentiating. Raise the exponent by one and divide by the new exponent; the derivative of your choice must be exactly

x^4

❌ Check the evaluation. With

F(x) = \tfrac{x^5}{5}

, the top value is

\tfrac{2^5}{5} = \tfrac{32}{5}

and the bottom value is

\tfrac{1^5}{5} = \tfrac{1}{5}

, which is not zero here; subtract it from the top value.

Show solution

Step 1: Antiderivative. By the power rule, $F(x) = \tfrac{x^5}{5}$ . Check: $F'(x) = \tfrac{5x^4}{5} = x^4$ . (The derivative of $4x^3$ is $12x^2$ , of $x^5$ is $5x^4$ , and of $\tfrac{x^5}{4}$ is $\tfrac{5}{4}x^4$ .)

Step 2: Evaluate.

\int_1^2 x^4\,dx = \frac{x^5}{5}\,\Big|_1^2 = \frac{2^5}{5} - \frac{1^5}{5} = \frac{32}{5} - \frac{1}{5} = \frac{31}{5}.

Problem 4 · A Fractional Power

Find the area under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ , that is, $\displaystyle\int_0^4 x^{1/2}\,dx$ .

✅ Correct! The power rule works for every exponent except

-1

, fractions included:

\tfrac{2}{3}x^{3/2}\big|_0^4 = \tfrac{2}{3}\cdot 8 = \tfrac{16}{3}

❌ Divide by the new exponent. You divided by the old exponent

\tfrac{1}{2}

. The new exponent is

\tfrac{1}{2} + 1 = \tfrac{3}{2}

❌ Divide, don't multiply.

\tfrac{3}{2}x^{3/2}

has derivative

\tfrac{9}{4}x^{1/2}

, not

x^{1/2}

. The antiderivative is

x^{3/2}

divided by

\tfrac{3}{2}

❌ Not quite.

4^{3/2} = 8

is the top value of

x^{3/2}

, but the antiderivative is

\tfrac{x^{3/2}}{3/2}

, so you still have to divide by

\tfrac{3}{2}

❌ Not quite. Write

\sqrt{x} = x^{1/2}

and apply the power rule.

Show solution

With $n = \tfrac{1}{2}$ , the power rule gives $F(x) = \dfrac{x^{3/2}}{3/2} = \tfrac{2}{3}x^{3/2}$ . Check: $F'(x) = \tfrac{2}{3}\cdot\tfrac{3}{2}x^{1/2} = x^{1/2}$ .

\int_0^4 x^{1/2}\,dx = \frac{2}{3}x^{3/2}\,\Big|_0^4 = \frac{2}{3}\cdot 4^{3/2} - 0 = \frac{2}{3}\cdot 8 = \frac{16}{3}.

Sanity check: the region lies inside the $4 \times 2$ rectangle (area $8$ ) and contains the triangle with vertices $(0,0)$ , $(4,0)$ , $(4,2)$ (area $4$ ), and $4 < \tfrac{16}{3} \approx 5.33 < 8$ .

Solved: 0 / 4

The Fundamental Theorem of Calculus: Area in One Line

IKey Formulas

IIArea Under fff Equals the Rise of FFF

IIIOne Hump of Sine

IVThe Power Rule on [0,1][0, 1][0,1]

VQuiz Questions

IIArea Under $f$ Equals the Rise of $F$

IVThe Power Rule on $[0, 1]$