Difference between revisions of "009B Sample Final 1, Problem 2"

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||So, we have
 
||So, we have
 
|-
 
|-
|<math>f(x)=\int_{-1}^{x} \sin(t^2)2t~dt=\int_{1}^{x^2} \sin(u)du=\left.-\cos(u)\right|_{1}^{x^2}=-\cos(x^2)+\cos(1)</math>.
+
|
 +
::<math>\begin{array}{rcl}
 +
f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
 +
&&\\
 +
& = & \displaystyle{\int_{1}^{x^2} \sin(u)~du}\\
 +
&&\\
 +
& = & \displaystyle{-\cos(u)\bigg|_{1}^{x^2}}\\
 +
&&\\
 +
& = & \displaystyle{-\cos(x^2)+\cos(1)}\\
 +
\end{array}</math>
 
|}
 
|}
  

Revision as of 12:34, 10 February 2016

We would like to evaluate

.

a) Compute .

b) Find .

c) State the fundamental theorem of calculus.

d) Use the fundamental theorem of calculus to compute without first computing the integral.

Foundations:  
Review -substitution

Solution:

(a)

Step 1:  
We proceed using -substitution. Let . Then, .
Since this is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation , we get and .
Step 2:  
So, we have

(b)

Step 1:  
From part (a), we have .
Step 2:  
If we take the derivative, we get .

(c)

Step 1:  
The Fundamental Theorem of Calculus has two parts.
The Fundamental Theorem of Calculus, Part 1
Let be continuous on and let .
Then, is a differentiable function on and .
Step 2:  
The Fundamental Theorem of Calculus, Part 2
Let be continuous on and let be any antiderivative of .
Then,

(d)

Step 1:  
By the Fundamental Theorem of Calculus, Part 1,
Final Answer:  
(a)
(b)
(c) The Fundamental Theorem of Calculus, Part 1
Let be continuous on and let .
Then, is a differentiable function on and .
The Fundamental Theorem of Calculus, Part 2
Let be continuous on and let be any antiderivative of .
Then, .
(d)

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