Evaluating Definite Integrals Using the Fundamental Theorem of Calculus

Hello. This course builds toward evaluating and testing improper integrals, then using the Beta and Gamma functions. Before an integral becomes “improper,” you need to be fully reliable with an ordinary, finite definite integral. This first lesson establishes that essential calculation tool: the Fundamental Theorem of Calculus.

By the end, you should be able to find an antiderivative, substitute the upper and lower limits correctly, and obtain the numerical value of a definite integral.

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1. Definite integrals and antiderivatives

A definite integral has lower and upper limits:

$${\int }_a^{b}f(x)dx$$

Here:

• $f(x)$ is the integrand.
• $x$ is the variable of integration.
• $a$ is the lower limit.
• $b$ is the upper limit.

For now, assume $f$ is continuous and finite on the full closed interval $[a,b]$. Such an integral is called a proper definite integral.

An antiderivative of $f(x)$ is a function $F(x)$ whose derivative is $f(x)$:

$$F^{\prime }(x)=f(x)$$

For example,

$$\frac{d}{dx}\left(x^3\right)=3x^2,$$

so an antiderivative of $3x^2$ is $x^3$.

The key fact is that integration and differentiation are inverse processes. The Fundamental Theorem of Calculus lets us use an antiderivative to calculate a definite integral exactly, without adding many small rectangles or using approximations.

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2. The Fundamental Theorem of Calculus

The evaluation form of the Fundamental Theorem of Calculus says:

If $F^{\prime }(x)=f(x)$ on $[a,b]$, then

$${\int }_a^{b}f(x)dx=F(b)-F(a).$$

Read this as:

1. Find an antiderivative $F(x)$ of $f(x)$.
2. Put the upper limit $b$ into $F$.
3. Put the lower limit $a$ into $F$.
4. Subtract:

$$\text{upper value}-\text{lower value}.$$

A compact notation is:

$${\int }_a^{b}f(x)dx={F(x)|}_a^b=F(b)-F(a).$$

The vertical bar does not mean “substitute both limits at the same time.” It is an instruction to calculate the value at the top limit and subtract the value at the bottom limit.

Why we usually omit $+C$

For an indefinite integral, we write:

$$\int f(x)dx=F(x)+C.$$

The constant $C$ is necessary because many functions have the same derivative. For example, both $x^2$ and $x^2+7$ differentiate to $2x$.

For a definite integral, however, the constant cancels:

$$[F(x)+C{]}_a^b=[F(b)+C]-[F(a)+C]=F(b)-F(a).$$

So when evaluating a definite integral, write an antiderivative without $+C$. This is not a shortcut that changes the answer; the constant genuinely has no effect.

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3. The reliable exam method

Use the following layout in every solution. It prevents the most common sign and substitution errors.

Given

$${\int }_a^{b}f(x)dx,$$

write:

$${\int }_a^{b}f(x)dx={F(x)|}_a^b=F(b)-F(a).$$

Then simplify carefully.

Worked example 1: A polynomial

Evaluate

$${\int }_{-1}^2(3x^2-4x+1)dx.$$

Step 1: Find an antiderivative.

$$\int (3x^2-4x+1)dx=x^3-2x^2+x.$$

Thus, take

$$F(x)=x^3-2x^2+x.$$

Step 2: Apply upper minus lower.

$${\int }_{-1}^2(3x^2-4x+1)dx={\left[x^3-2x^2+x\right]}_{-1}^2.$$ $$=\left(2^3-2(2^2)+2\right)-\left((-1{)}^3-2(-1{)}^2+(-1)\right).$$

Step 3: Simplify each bracket separately.

$$=(8-8+2)-(-1-2-1).$$ $$=2-(-4)=6.$$

Therefore,

$$\boxed{{\int }_{-1}^2(3x^2-4x+1)dx=6.}$$

The lower limit is negative, so parentheses are essential. Without them, it is easy to lose the outer subtraction sign.

A useful antiderivative rule

For any real number $n\ne -1$,

$$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C.$$

For a definite integral, use the same rule but omit $+C$.

For example,

$$\int 5x^{4}dx=x^5,$$

because

$$\frac{d}{dx}(x^5)=5x^4.$$

Be careful: the power rule does not apply when $n=-1$. That special case is

$$\int \frac{1}{x}dx=\ln |x|+C.$$

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4. Different forms, same theorem

The Fundamental Theorem works whenever you can find an antiderivative and the function is continuous on the interval. The integrand may be a constant, polynomial, trigonometric function, exponential function, logarithmic derivative, or a fractional power.

Worked example 2: A square-root expression

Evaluate

$${\int }_1^4\frac{2}{\sqrt{x}}dx.$$

First rewrite the radical as a power:

$$\frac{2}{\sqrt{x}}=2x^{-1/2}.$$

Now integrate using the power rule:

$$\int 2x^{-1/2}dx=2\left(\frac{x^{1/2}}{1/2}\right)=4x^{1/2}.$$

Apply the limits:

$${\int }_1^4\frac{2}{\sqrt{x}}dx={\left[4\sqrt{x}\right]}_1^4.$$ $$=4\sqrt{4}-4\sqrt{1}.$$ $$=8-4=4.$$

Hence,

$$\boxed{{\int }_1^4\frac{2}{\sqrt{x}}dx=4.}$$

Worked example 3: The logarithm case

Evaluate

$${\int }_1^e\frac{5}{x}dx.$$

An antiderivative of $\frac{5}{x}$ is $5\ln x$ on this positive interval. Therefore,

$${\int }_1^e\frac{5}{x}dx={\left[5\ln x\right]}_1^e.$$ $$=5\ln e-5\ln 1.$$

Using $\ln e=1$ and $\ln 1=0$,

$$\boxed{{\int }_1^e\frac{5}{x}dx=5.}$$

This example will matter later: although $\frac{1}{x}$ is perfectly continuous on $[1,e]$, it is not defined at $x=0$. If an integration interval reaches $0$, the integral needs a different treatment and may be improper.

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5. What a definite integral means

A definite integral gives net signed area between the graph of $y=f(x)$ and the $x$-axis.

• Where $f(x)>0$, the integral contributes positively.
• Where $f(x)<0$, the integral contributes negatively.
• The final value may be negative, zero, or positive.

For instance, if a function has equal positive and negative signed contributions over an interval, its definite integral can be zero even though there is visible geometric area above and below the axis.

A simple symmetry result follows. If $f$ is an odd function, meaning

$$f(-x)=-f(x),$$

then over a symmetric interval,

$${\int }_{-a}^{a}f(x)dx=0.$$

For example,

$${\int }_{-2}^2{x}^{3}dx=0.$$

This does not say there is no area. It says that the positive and negative signed areas cancel.

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6. Common errors and how to avoid them

1. Forgetting to subtract the lower-limit value

Incorrect idea:

$$[F(x){]}_a^b=F(b).$$

Correct rule:

$$[F(x){]}_a^b=F(b)-F(a).$$

2. Losing signs at a negative lower limit

If the lower limit is $-1$, write:

$$F(2)-F(-1),$$

not just $F(2)-F-1$. Substitute using parentheses:

$$(-1{)}^3,(-1{)}^2.$$

3. Adding $+C$ to a definite integral

It will cancel, but it makes your work longer and may create sign mistakes. Write the antiderivative without $+C$.

4. Applying the power rule incorrectly

For

$$\int x^{n}dx,$$

increase the exponent by $1$, then divide by the new exponent:

$$\int x^{n}dx=\frac{x^{n+1}}{n+1}.$$

For example,

$$\int x^{3}dx=\frac{x^4}{4},$$

not $x^4$.

5. Confusing $\frac{1}{x}$ with a usual power-rule example

The formula fails for $x^{-1}$. Remember:

$$\int \frac{1}{x}dx=\ln |x|+C.$$

6. Forgetting that the theorem has conditions

The direct formula

$${\int }_a^{b}f(x)dx=F(b)-F(a)$$

is used immediately for functions that are continuous and finite on the whole interval. In later lessons, intervals extending indefinitely or functions becoming unbounded require limits first. Those are the improper integrals this course focuses on.

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A compact solution template

For exam work, use this format:

$${\int }_a^{b}f(x)dx$$ $$={\left[F(x)\right]}_a^b$$ $$=F(b)-F(a)$$ $$=\text{final numerical value}.$$

Before moving on, check:

• Did I find an antiderivative correctly?
• Did I substitute the upper limit first?
• Did I subtract the complete lower-limit expression?
• Did I use parentheses around negative values?
• Is the integrand finite and continuous on the stated interval?

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You have now established the calculation method underlying the whole unit: a proper definite integral is evaluated by taking an antiderivative and computing upper value minus lower value. The constant of integration cancels, and careful parentheses protect you from sign errors.

Next, you will review the limits involving powers, logarithms, and exponentials that are needed when the endpoints or integrand values are no longer finite.