Velocity due to gravity can be easily calculated by the formula: v = gt, where g is the acceleration due to gravity (9.8m/s^{2}) 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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 KhanAcademy  The Indefinite Integral
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


There are rules for finding the antiderivatives of simple power functions such as f(x) = x^{2}. 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 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


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 \begin{align*}\int_{1}^{2} x^2 dx.\end{align*}
Solution
This integral tells us to evaluate the area under the curve f(x) = x^{2}, 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) = x^{2}. It turns out to be F(x) = (1/3)x^{3} + C, where C is a constant of integration. How can we get this? Think about the functions that will have derivatives of x^{2}. 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,
\begin{align*}\int_{a}^{b} f(x) dx\end{align*}  \begin{align*}= F(x)^b_a\end{align*}  

\begin{align*}\int_{1}^{2} x^2 dx\end{align*}  \begin{align*}= \left [\frac{1} {3} x^3 + C\right ]^2_1\end{align*}  
\begin{align*}= \left [\frac{1} {3} (2)^3 + C\right ]  \left [\frac{1} {3} (1)^3 + C\right ]\end{align*}  
\begin{align*}= \left [\frac{8} {3} + C\right ]  \left [\frac{1} {3} + C\right ]\end{align*}  
\begin{align*}= \frac{7} {3} + C  C\end{align*}  
\begin{align*}= \frac{7} {3}\end{align*} 
So the area under the curve is (7/3) units^{2}.
Example B
Evaluate \begin{align*}\int x^3dx.\end{align*}
Solution
Since \begin{align*}\int x^n dx = \frac{1} {n + 1}x^{n + 1} + C\end{align*}, we have
\begin{align*}\int x^3 dx\end{align*}  \begin{align*}= \frac{1} {3 + 1}x^{3 + 1} + C\end{align*}  

\begin{align*}= \frac{1} {4}x^4 + C\end{align*} 
 To check our answer we can take the derivative of \begin{align*} \frac{1} {4}x^4 + C\end{align*} and verify that it is \begin{align*}\,\! x^3\end{align*}, the original function in our integral.
Example C
Evaluate \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:



 \begin{align*}\int 5x^2 dx = 5 \int x^2 dx\end{align*}


 Then we can integrate:
\begin{align*}= 5 \cdot \frac{1} {2 + 1} x^{2 + 1} + C\end{align*}  

\begin{align*}= \frac{5} {3} x^3 + C\end{align*} 
 Again, if we wanted to check our work we could take the derivative of \begin{align*}\frac{5} {3} x^3 + C\end{align*} and verify that we get \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 xaxis) 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 \begin{align*}\int (3x^3  4x^2 + 2)dx.\end{align*}
2) Evaluate \begin{align*}\int_{2}^{5} \sqrt{x} dx.\end{align*}
3)Use the fundamental theorem of calculus to solve: \begin{align*}\int_{4}^{6} \frac{dx}{x}\end{align*}
4)Use the fundamental theorem of calculus to solve:\begin{align*}\int_{2p}^{2p} 3cos(x) dx =\end{align*}
5)Use the 2nd fundamental theorem of calculus to solve:\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:

\begin{align*}\int (3x^3  4x^2 + 2)dx =\end{align*}
 \begin{align*}3 \left( \int x^3 dx \right)  4 \left( \int x^2 dx \right) + \left( \int 2dx \right)\end{align*}
\begin{align*}\to 3 \cdot \frac{1} {4} x^4  4 \cdot \frac{1} {3}x^3 + 2x + C\end{align*} \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 \begin{align*}y = \sqrt{x}\end{align*} from x = 2 to x = 3, shown in the figure below.
\begin{align*}\int_{2}^{5} \sqrt{x} dx\end{align*}  \begin{align*}= \int_{2}^{5} x^{1/2} dx\end{align*}  

\begin{align*}= \left [\frac{1} {\frac{1} {2} + 1} x^{1/2 + 1}\right ]^5_2\end{align*}  
\begin{align*}= \left [\frac{1} {3/2} x^{3/2}\right ]^5_2\end{align*}  
\begin{align*}= \frac{2} {3} \left [x^{3/2}\right ]^5_2\end{align*}  
\begin{align*}= \frac{2} {3}\left [5^{3/2}  2^{3/2}\right ]\end{align*}  
\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) = \begin{align*}\frac {1}{x}\end{align*}
 Thus, we apply the fundamental theorem of calculus:
 \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:
 \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 \begin{align*}A'(\frac{3p}{4})\end{align*}
 The second theorem states:\begin{align*}\int_{3}^{x}cot^3 (t) dt\end{align*} is the antiderivative of \begin{align*}cot^3(x)\end{align*}
 So, \begin{align*}A'(x) = cot^3(x)\end{align*}
 Substituting in \begin{align*}x = \frac{3p}{4}\end{align*} we get an answer of \begin{align*}A'{4} = 1\end{align*}
Practice
Evaluate the Integral:
 Evaluate the integral \begin{align*}\int_{0}^{3} 5xdx\end{align*}
 Evaluate the integral \begin{align*}\int_{0}^{1} x^4dx\end{align*}
 Evaluate the integral \begin{align*}\int_{1}^{4} (x  3)dx\end{align*}
Find the Integral:
 Find the integral of (x + 1)(2x  3) from 1 to 2.
 Find the integral of \begin{align*}\sqrt{x}\end{align*} from 0 to 9.
 Find the integral of \begin{align*}\int_{1}^{0}  3 dx =\end{align*}
 Find the integral of \begin{align*}\int_{1}^{3} dx =\end{align*}
 Find the integral of \begin{align*}\int_{p}^{\frac{p}{2}}  4cos(x) dx =\end{align*}
 Find the integral of \begin{align*}\int_{0}^{2} dx =\end{align*}
 Find the integral of \begin{align*}\int_{2}^{7} \frac{dx}{x}\end{align*}
 Find the integral of \begin{align*}\int_{2}^{0}x + 5 dx =\end{align*}
 Find the integral of \begin{align*}\int_{p}^{\frac{3p}{2}}6sin(x) dx =\end{align*}
 Find the integral of \begin{align*}\int_{6}^{7} \frac{dx}{x}\end{align*}
Challenge yourself:
 Sketch y = x^{3} 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.
 Evaluate the integral \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:
 \begin{align*}A(x) = \int_{2}^{x} tan^3(t) dt\end{align*}
 Find:\begin{align*}\frac{d}{dx} \int_{1}^{x} csc^2(t) dt\end{align*}
 Find:\begin{align*}\frac{d}{dx} \int_{2}^{x^2} sec(t) dt\end{align*}