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# 8.13: Fundamental Theorem of Calculus

Difficulty Level: At Grade Created by: CK-12
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Velocity due to gravity can be easily calculated by the formula: v = gt, where g is the acceleration due to gravity (9.8m/s2) and t is time in seconds. In fact, a decent approximation can be calculated in your head easily by rounding 9.8 to 10 so you can just add a decimal place to the time.

Using this function for velocity, how could you find a function that represented the position of the object after a given time? What about a function that represented the instantaneous acceleration of the object at a given time?

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### Guidance

If you think that evaluating areas under curves is a tedious process you are probably right. Fortunately, there is an easier method. In this section, we shall give a general method of evaluating definite integrals (area under the curve) by using antiderivatives.

Definition: The Antiderivative
If F ' (x) = f (x), then F '(x) is said to be the antiderivative of f(x).

There are rules for finding the antiderivatives of simple power functions such as f(x) = x2. As you read through them, try to think about why they make sense, keeping in mind that differentiation reverses integration.

Rules of Finding the Antiderivatives of Power Functions
• The Power Rule
xndx=1n+1xn+1+C\begin{align*}\int x^n dx = \frac{1} {n + 1} x^{n + 1} + C\end{align*}
where C is constant of integration and n is a rational number not equal to -1.
• A Constant Multiple of a Function Rule
kxndx=kxndx=k1n+1xn+1+C\begin{align*}\int k x^n dx = k \int x^n dx = k \cdot \frac{1} {n + 1} x^{n + 1} + C\end{align*}
where k is a constant.
• Sum and Difference Rule
[f(x)±g(x)]dx=f(x)dx±g(x)dx\begin{align*}\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx\end{align*}
• The Constant rule
kdx=kx+C\begin{align*}\int k \cdot dx = kx + C\end{align*}
where k is a constant. (Notice that this rule comes as a result of the power rule above.)

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus makes the relationship between derivatives and integrals clear. Integration performed on a function can be reversed by differentiation.

The Fundamental Theorem of Calculus
If a function f(x) is defined over the interval [a, b] and if F(x) is the antidervative of f on [a, b], then
baf(x)dx=F(x)|ba\begin{align*}\int_{a}^{b} f(x) dx = F(x)|^b_a\end{align*}
=F(b)F(a)\begin{align*}= F(b) - F(a)\end{align*}

We can use the relationship between differentiation and integration outlined in the Fundamental Theorem of Calculus to compute definite integrals more quickly.

#### Example A

Evaluate 21x2dx.\begin{align*}\int_{1}^{2} x^2 dx.\end{align*}

Solution

This integral tells us to evaluate the area under the curve f(x) = x2, which is a parabola over the interval [1, 2], as shown in the figure below.

To compute the integral according to the Fundamental Theorem of Calculus, we need to find the antiderivative of f(x) = x2. It turns out to be F(x) = (1/3)x3 + C, where C is a constant of integration. How can we get this? Think about the functions that will have derivatives of x2. Take the derivative of F(x) to check that we have found such a function. (For more specific rules, see the box after this example). Substituting into the fundamental theorem,

baf(x)dx\begin{align*}\int_{a}^{b} f(x) dx\end{align*} =F(x)|ba\begin{align*}= F(x)|^b_a\end{align*}
21x2dx\begin{align*}\int_{1}^{2} x^2 dx\end{align*} =[13x3+C]21\begin{align*}= \left [\frac{1} {3} x^3 + C\right ]^2_1\end{align*}
=[13(2)3+C][13(1)3+C]\begin{align*}= \left [\frac{1} {3} (2)^3 + C\right ] - \left [\frac{1} {3} (1)^3 + C\right ]\end{align*}
=[83+C][13+C]\begin{align*}= \left [\frac{8} {3} + C\right ] - \left [\frac{1} {3} + C\right ]\end{align*}
=73+CC\begin{align*}= \frac{7} {3} + C - C\end{align*}
=73\begin{align*}= \frac{7} {3}\end{align*}

So the area under the curve is (7/3) units2.

#### Example B

Evaluate x3dx.\begin{align*}\int x^3dx.\end{align*}

Solution

Since xndx=1n+1xn+1+C\begin{align*}\int x^n dx = \frac{1} {n + 1}x^{n + 1} + C\end{align*}, we have

x3dx\begin{align*}\int x^3 dx\end{align*} =13+1x3+1+C\begin{align*}= \frac{1} {3 + 1}x^{3 + 1} + C\end{align*}
=14x4+C\begin{align*}= \frac{1} {4}x^4 + C\end{align*}
To check our answer we can take the derivative of 14x4+C\begin{align*} \frac{1} {4}x^4 + C\end{align*} and verify that it is x3\begin{align*}\,\! x^3\end{align*}, the original function in our integral.

#### Example C

Evaluate 5x2dx.\begin{align*}\int 5x^2 dx.\end{align*}

Solution

Using the constant multiple of a power rule, the coefficient 5 can be removed outside the integral:

5x2dx=5x2dx\begin{align*}\int 5x^2 dx = 5 \int x^2 dx\end{align*}
Then we can integrate:
=512+1x2+1+C\begin{align*}= 5 \cdot \frac{1} {2 + 1} x^{2 + 1} + C\end{align*}
=53x3+C\begin{align*}= \frac{5} {3} x^3 + C\end{align*}
Again, if we wanted to check our work we could take the derivative of 53x3+C\begin{align*}\frac{5} {3} x^3 + C\end{align*} and verify that we get 5x2\begin{align*}\,\! 5x^2\end{align*}

### Vocabulary

The Fundamental Theorem of Calculus demonstrates that integration performed on a function can be reversed by differentiation.

Integrals allow for the calculation of the area between a line (such as the x-axis) and a curve, or of the area between two curves. Since the area is generally given in square units, it is technically only an approximation, but can be an effectively infinitely close one!

The antiderivative has much the same relationship to a function that a square root has to a constant. The antiderivative of a function is the function whose derivative is the function you want the antiderivative of.

### Guided Practice

Questions

1) Evaluate (3x34x2+2)dx.\begin{align*}\int (3x^3 - 4x^2 + 2)dx.\end{align*}

2) Evaluate 52xdx.\begin{align*}\int_{2}^{5} \sqrt{x} dx.\end{align*}

3)Use the fundamental theorem of calculus to solve: 64dxx\begin{align*}\int_{4}^{6} \frac{dx}{x}\end{align*}

4)Use the fundamental theorem of calculus to solve:2p2p3cos(x)dx=\begin{align*}\int_{-2p}^{2p} 3cos(x) dx =\end{align*}

5)Use the 2nd fundamental theorem of calculus to solve:A(x)=x3cot3(t)dt\begin{align*}A(x) = \int_{3}^{x} cot^3 (t) dt\end{align*}

Solutions

1) Using the sum and difference rule we can separate our integral into three integrals:

(3x34x2+2)dx=\begin{align*}\int (3x^3 - 4x^2 + 2)dx =\end{align*}
3(x3dx)4(x2dx)+(2dx)\begin{align*}3 \left( \int x^3 dx \right) - 4 \left( \int x^2 dx \right) + \left( \int 2dx \right)\end{align*}

314x4413x3+2x+C\begin{align*}\to 3 \cdot \frac{1} {4} x^4 - 4 \cdot \frac{1} {3}x^3 + 2x + C\end{align*} 34x443x3+2x+C\begin{align*}\to \frac{3} {4} x^4 - \frac{4} {3} x^3 + 2x + C\end{align*}

2) The evaluation of this integral represents calculating the area under the curve y=x\begin{align*}y = \sqrt{x}\end{align*} from x = -2 to x = 3, shown in the figure below.

52xdx\begin{align*}\int_{2}^{5} \sqrt{x} dx\end{align*} =52x1/2dx\begin{align*}= \int_{2}^{5} x^{1/2} dx\end{align*}
=[112+1x1/2+1]52\begin{align*}= \left [\frac{1} {\frac{1} {2} + 1} x^{1/2 + 1}\right ]^5_2\end{align*}
=[13/2x3/2]52\begin{align*}= \left [\frac{1} {3/2} x^{3/2}\right ]^5_2\end{align*}
=23[x3/2]52\begin{align*}= \frac{2} {3} \left [x^{3/2}\right ]^5_2\end{align*}
=23[53/223/2]\begin{align*}= \frac{2} {3}\left [5^{3/2} - 2^{3/2}\right ]\end{align*}
=5.57\begin{align*}= 5.57\end{align*}
So the area under the curve is 5.57.

3) Given what we know, that if F(x) = ln x, then F'(x) = 1x\begin{align*}\frac {1}{x}\end{align*}

Thus, we apply the fundamental theorem of calculus:
46dxx=lnx|46\begin{align*}\int-{4}^{6} \frac{dx}{x} = ln x |-{4}{6}\end{align*}
= F(6) - F(4) = [ln(6)] - [ln(4)] = 0.4055

4) Given what we know, that if F(x) = 3sin(x), then F'(x) = 3cos(x)

So we apply the fundamental theorem of calculus:
2p2p3cosdx=3sin(x)|2p2p\begin{align*}\int_{-2p}^{2p} 3cos dx = 3sin(x) | _{-2p}^{2p}\end{align*}
= F(8) - F(0) = [3sin(2p)] - [3sin(-2p)] = 1 - 0 = 0

5) Find A(3p4)\begin{align*}A'(\frac{3p}{4})\end{align*}

The second theorem states:x3cot3(t)dt\begin{align*}\int_{3}^{x}cot^3 (t) dt\end{align*} is the anti-derivative of cot3(x)\begin{align*}cot^3(x)\end{align*}
So, A(x)=cot3(x)\begin{align*}A'(x) = cot^3(x)\end{align*}
Substituting in x=3p4\begin{align*}x = \frac{3p}{4}\end{align*} we get an answer of A4=1\begin{align*}A'{4} = -1\end{align*}

### Practice

Evaluate the Integral:

1. Evaluate the integral 305xdx\begin{align*}\int_{0}^{3} 5xdx\end{align*}
2. Evaluate the integral 10x4dx\begin{align*}\int_{0}^{1} x^4dx\end{align*}
3. Evaluate the integral 41(x3)dx\begin{align*}\int_{1}^{4} (x - 3)dx\end{align*}

Find the Integral:

1. Find the integral of (x + 1)(2x - 3) from -1 to 2.
2. Find the integral of x\begin{align*}\sqrt{x}\end{align*} from 0 to 9.
3. Find the integral of 013dx=\begin{align*}\int_{-1}^{0} - 3 dx =\end{align*}
4. Find the integral of 31dx=\begin{align*}\int_{-1}^{3} dx =\end{align*}
5. Find the integral of p2p4cos(x)dx=\begin{align*}\int_{-p}^{\frac{p}{2}} - 4cos(x) dx =\end{align*}
6. Find the integral of 20dx=\begin{align*}\int_{0}^{2} -dx =\end{align*}
7. Find the integral of 72dxx\begin{align*}\int_{2}^{7} \frac{dx}{x}\end{align*}
8. Find the integral of 02x+5dx=\begin{align*}\int_{-2}^{0}x + 5 dx =\end{align*}
9. Find the integral of 3p2p6sin(x)dx=\begin{align*}\int_{-p}^{\frac{3p}{2}}6sin(x) dx =\end{align*}
10. Find the integral of 76dxx\begin{align*}\int_{6}^{7} \frac{dx}{x}\end{align*}

Challenge yourself:

1. Sketch y = x3 and y = x on the same coordinate system and then find the area of the region enclosed between them (a) in the first quadrant and (b) in the first and third quadrants.
2. Evaluate the integral RR(πR2πx2)dx\begin{align*}\int_{-R}^{R} (\pi R^2 - \pi x^2) dx\end{align*} where R is a constant.

Apply the second theorem of calculus:

1. A(x)=x2tan3(t)dt\begin{align*}A(x) = \int_{2}^{x} tan^3(t) dt\end{align*}
2. Find:ddxx1csc2(t)dt\begin{align*}\frac{d}{dx} \int_{1}^{x} csc^2(t) dt\end{align*}
3. Find:ddxx22sec(t)dt\begin{align*}\frac{d}{dx} \int_{-2}^{x^2} sec(t) dt\end{align*}

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### Vocabulary Language: English

antiderivative

An antiderivative is a function that reverses a derivative. Function A is the antiderivative of function B if function B is the derivative of function A.

derivative

The derivative of a function is the slope of the line tangent to the function at a given point on the graph. Notations for derivative include $f'(x)$, $\frac{dy}{dx}$, $y'$, $\frac{df}{dx}$ and \frac{df(x)}{dx}.

fundamental theorem of calculus

The fundamental theorem of calculus demonstrates that integration performed on a function can be reversed by differentiation.

integral

An integral is used to calculate the area under a curve or the area between two curves.

theorem

A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven.

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Date Created:
Nov 14, 2012
May 26, 2016
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