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CHAPTER 4. INTEGRALS 105 4.4 Using Geometry for Definite Integrals Graph the integrands and use geometry to evaluate the definite integrals. 911. 4 2 x 2 +3 dx 912. 3 3 9 x 2 dx 913. 1 2 |x| dx 914. 1 1 (2 |x|) dx 915. b 0 x dx where b> 0 916. b a 2x dx where 0 <a<b 917. Suppose f and g are continuous and that 2 1 f (x) dx = 4, 5 1 f (x) dx =6, 5 1 g(x) dx =8. Evaluate the following definite integrals. a) 2 2 g(x) dx b) 1 5 g(x) dx c) 2 1 3f (x) dx d) 5 2 f (x) dx e) 5 1 [f (x) g(x)] dx f) 5 1 [4f (x) g(x)] dx 918. Suppose that 0 3 g(t) dt = 2. Find the following. a) 3 0 g(t) dt b) 0 3 g(u) du c) 0 3 g(x) dx d) 0 3 g(θ) 2 dθ 919. A particle moves along the x-axis so that at any time t 0 its acceleration is given by a(t) = 18 2t. At time t = 1 the velocity of the particle is 36 meters per second and its position is x = 21. a) Find the velocity function and the position function for t 0. b) What is the position of the particle when it is farthest to the right? When you feel how depressingly Slowly you climb, It’s well to remember That things take time. —Piet Hein
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4.4 Using Geometry for Definite Integralsmcalc.weebly.com/uploads/2/2/5/0/22502010/calculus_ap_-_geometr… · INTEGRALS 105 4.4 Using Geometry for Definite Integrals Graph the

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Page 1: 4.4 Using Geometry for Definite Integralsmcalc.weebly.com/uploads/2/2/5/0/22502010/calculus_ap_-_geometr… · INTEGRALS 105 4.4 Using Geometry for Definite Integrals Graph the

CHAPTER 4. INTEGRALS 105

4.4 Using Geometry for Definite IntegralsGraph the integrands and use geometry to evaluate the definite integrals.

911.! 4

−2

"x

2+ 3#

dx

912.! 3

−3

$

9 − x2 dx

913.! 1

−2|x| dx

914.! 1

−1(2 − |x|) dx

915.! b

0x dx where b > 0

916.! b

a2x dx where 0 < a < b

917. Suppose f and g are continuous and that

! 2

1f(x) dx = −4,

! 5

1f(x) dx = 6,

! 5

1g(x) dx = 8.

Evaluate the following definite integrals.

a)

! 2

2g(x) dx

b)

! 1

5g(x) dx

c)

! 2

13f(x) dx

d)

! 5

2f(x) dx

e)

! 5

1[f(x) − g(x)] dx

f)

! 5

1[4f(x) − g(x)] dx

918. Suppose that

! 0

−3g(t) dt =

√2. Find the following.

a)

! −3

0g(t) dt b)

! 0

−3g(u) du c)

! 0

−3−g(x) dx d)

! 0

−3

g(θ)√2

919. A particle moves along the x-axis so that at any time t ≥ 0 its acceleration is given bya(t) = 18−2t. At time t = 1 the velocity of the particle is 36 meters per second and its positionis x = 21.

a) Find the velocity function and the position function for t ≥ 0.

b) What is the position of the particle when it is farthest to the right?

When you feel how depressinglySlowly you climb,It’s well to rememberThat things take time.

—Piet Hein