Calculus 42
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Antiderivativves
5.6
Find all possible functions f with the given derivative.
1. f ′ ( x ) = 2 x 2. f ′ ( x ) = 5
′ ( x ) = 3x 2 + x − 4 3. f
The process of finding the original function from the derivative is so important that it has a name:
1. f ′ ( x ) = 2 x
Find the function f ( x ) whose derivative is sin ( x ) and whose graph passes through ( 0, 2 ) .
d cos ( x ) = − sin ( x ) dx
so:
∴ f ( x ) = − cos ( x ) + C
f ( x ) = − cos ( x ) + 3
→
Page 291 #1-21 all
You will hear much more about antiderivatives in the future. This section is just an introduction.
→
Find all possible functions f with the given derivative.
Antiderivative A function F ( x ) is an antiderivative of a function f ( x ) of finding an antiderivative is antidifferentiation. if F ′ ( x ) = f ( x ) for all x in the domain of f. The process
d − cos ( x ) =wenku.baidu.comsin ( x ) dx
2 = − cos ( 0 ) + C
f ( x ) could be − cos ( x ) or could vary by some constant C .
Find the function f ( x ) whose derivative is sin ( x ) and whose graph passes through ( 0, 2 ) .
d cos ( x ) = − sin ( x ) dx
so:
∴ f ( x ) = − cos ( x ) + C
d − cos ( x ) = sin ( x ) dx
2 = − cos ( 0 ) + C
2 = −1 + C 3=C
Notice that we had to have initial values to determine the value of C.
5.6
Find all possible functions f with the given derivative.
1. f ′ ( x ) = 2 x 2. f ′ ( x ) = 5
′ ( x ) = 3x 2 + x − 4 3. f
The process of finding the original function from the derivative is so important that it has a name:
1. f ′ ( x ) = 2 x
Find the function f ( x ) whose derivative is sin ( x ) and whose graph passes through ( 0, 2 ) .
d cos ( x ) = − sin ( x ) dx
so:
∴ f ( x ) = − cos ( x ) + C
f ( x ) = − cos ( x ) + 3
→
Page 291 #1-21 all
You will hear much more about antiderivatives in the future. This section is just an introduction.
→
Find all possible functions f with the given derivative.
Antiderivative A function F ( x ) is an antiderivative of a function f ( x ) of finding an antiderivative is antidifferentiation. if F ′ ( x ) = f ( x ) for all x in the domain of f. The process
d − cos ( x ) =wenku.baidu.comsin ( x ) dx
2 = − cos ( 0 ) + C
f ( x ) could be − cos ( x ) or could vary by some constant C .
Find the function f ( x ) whose derivative is sin ( x ) and whose graph passes through ( 0, 2 ) .
d cos ( x ) = − sin ( x ) dx
so:
∴ f ( x ) = − cos ( x ) + C
d − cos ( x ) = sin ( x ) dx
2 = − cos ( 0 ) + C
2 = −1 + C 3=C
Notice that we had to have initial values to determine the value of C.