Aim:%%What%is%the%fundamental%theoremof%calculus...
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Aim: What is the fundamental theorem of calculus(#2)? Get Ready: 1. 2. 3. 4. I. Graphical: Analytical: ( x 2 + x ) dx 0 1 ∫ 3x 2 dx 1 3 ∫ −4 x 3 dx 1 5 ∫ x 3 2 dx 0 5 ∫ f ( x ) = sin( x ) F( x ) = − cos( x ) sin( x 0 π ∫ )dx sin( x 0 2 π ∫ )dx
Transcript of Aim:%%What%is%the%fundamental%theoremof%calculus...
Aim: What is the fundamental theorem of calculus(#2)? Get Ready:
1. 2.
3. 4. I. Graphical: Analytical:
(x2 + x )dx0
1
∫ 3x2 dx1
3
∫
−4x3 dx1
5
∫ x32 dx
0
5
∫
f (x) = sin(x) F(x) = −cos(x)
sin(x0
π
∫ )dx
sin(x0
2π
∫ )dx
Aim: What is the fundamental theorem of calculus(#2)? II. Trigonometric Review of Finding Exact Values of Trig Functions Process: 1. Identify the quadrant based upon the rotation 2. Find Reference Angle 3. Find sign (+/-‐) of trig function based upon the quadrant 4. Express as a function of a positive acute angle 5. Find Exact Value A. Quadrantals: Key Ideas: Sine Cosine Tangent 1)
2) 3)
4) B. Find the value of the sine, cosine, and tangent of the following rotations.
1) 2) 3) 4) 5)
C. Use Hand Trick to Develop sin-‐cos-‐tan values
1) 2)
3) 4)
sin0° = cos0° = tan0° =
sin(π2) cos(π
2) = tan(π
2) =
sin(π ) = cos(π ) = tan(π ) =
sin(3π2) = cos(3π
2) = tan(3π
2) =
5π2 3π
7π2
4π 5π
θ 0° 30° 45° 60° 90°
sinθ
cosθ
tanθ
sin(3π4) cos(2π
3)
tan(5π6) sin(7π
6)
360°
270°
180°
90°
0°
Aim: What is the fundamental theorem of calculus(#2)?
5) 6)
7) 8)
9) 10)
11) 12)
13) 14)
15) 16)
17) 18)
19) 20)
21) 22)
23) 24)
cos(5π4) tan(− 5π
4)
sin(− π4) sin(3π
2)
cos(3π2)
cos0
cosπcos(5π
3)
cos(− π6) tan(5π
4)
tan(3π4) tan(5π
3)
tan(− 5π3)
tanπ
sin(π2) cos(11π
6)
tan(5π4) sin(5π
6)
tan(3π2) sin(0)
Aim: What is the fundamental theorem of calculus(#2)? III. Finding Definite Integral with Trig
1.
2.
3.
4.
5.
6.
7.
8.
f (x)dxa
b
∫ = F(b)− F(a)
sin xdx0
3π2
∫
sin xdxπ
2π
∫
sin xdx0
2π
∫
cos xdxπ2
3π2∫
sin xdx−π2
π2∫
− cos xdx−π
π
∫
cos xdx−π2
π2
∫ =
(2+ sin(x0
π
∫ ))dx
Aim: What is the fundamental theorem of calculus(#2)?
9.
10.
11.
12.
13.
14.
15.
16.
(2x − sin x)dx0
π
∫ =
(3sin x − 2cos x)dx0
π2
∫ =
sec x tan xdx0
π4∫
sec2 xdx0
π6
∫
(4x + sec2 x)dx0
π4
∫ =
(x − sec2 x)dx0
π4
∫ =
csc2 (x)π4
3π4
∫ dx
−cot(x)csc(x)π2
3π4
∫ dx
