Solve the problem. a) Suppose that ∫−9−6g(t)dt=6. Find ∫−9−6g(x)6dx and ∫−6−9−g(t)dt b) Suppose that h is continuous and that ∫−33h(x)dx=4 and ∫36h(x)dx=−12. Find ∫−36h(t)dt and ∫6−3h(t)dt

Step 1 a) ∫−9−6g(t)dt=6 ⇒∫−9−6g(x)6dx=11 ⇒16∫−9−6g(x)dx=16x6=1 Step 2 ⇒∫−6−9−g(t)dt=? {∵∫abf(x)dx=−∫baf(x)dx} (−)∫−6−9g(t)dt=(−)x(−1)∫−9−6g(t)dt=(+)∫−9−6g(t)dt=+6 {∫−9−6g(x)d6dx=1+∫−6−9−g(t)dt}=+6

b) ∫−33h(x)dx=4, ∫36h(x)dx=−12, ∫−36h(t)dt ∵∫−36h(t)dt=∫−33h(x)dx+∫36h(x)dx {∫−36h(t)dt=−8+∫6−3h(t)dt=+8}

Step 1 a) ∫−9−6g(t)dt=6 ⇒∫−9−6g(x)6dx=11 ⇒16∫−9−6g(x)dx=16x6=1 Step 2 ⇒∫−6−9−g(t)dt=? {∵∫abf(x)dx=−∫baf(x)dx} (−)∫−6−9g(t)dt=(−)x(−1)∫−9−6g(t)dt=(+)∫−9−6g(t)dt=+6 {∫−9−6g(x)d6dx=1+∫−6−9−g(t)dt}=+6

b) ∫−33h(x)dx=4, ∫36h(x)dx=−12, ∫−36h(t)dt ∵∫−36h(t)dt=∫−33h(x)dx+∫36h(x)dx {∫−36h(t)dt=−8+∫6−3h(t)dt=+8}

Answered: "Solve the problem. a) Suppose that ∫−9−6g(t)dt=6. Find..."

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Osvaldo Apodaca

Answered question

2021-12-13

Solve the problem.
a) Suppose that

\int_{- 9}^{- 6} g (t) d t = 6

. Find

\int_{- 9}^{- 6} \frac{g (x)}{6} d x

and

\int_{- 6}^{- 9} - g (t) d t

b) Suppose that h is continuous and that

\int_{- 3}^{3} h (x) d x = 4

and

\int_{3}^{6} h (x) d x = - 12

. Find

\int_{- 3}^{6} h (t) d t

and

\int_{6}^{- 3} h (t) d t

Answer & Explanation

puhnut1m

Beginner2021-12-14Added 33 answers

Step 1
a)

\int_{- 9}^{- 6} g (t) d t = 6

\Rightarrow \int_{- 9}^{- 6} \frac{g (x)}{6} d x = 11

\Rightarrow \frac{1}{6} \int_{- 9}^{- 6} g (x) d x = \frac{1}{6} x^{6} = 1

Step 2

\Rightarrow \int_{- 6}^{- 9} - g (t) d t = ?

{∵ \int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x}

(-) \int_{- 6}^{- 9} g (t) d t = (-) x (- 1) \int_{- 9}^{- 6} g (t) d t = (+) \int_{- 9}^{- 6} g (t) d t = + 6

{\int_{- 9}^{- 6} \frac{g (x) d}{6} d x = 1 + \int_{- 6}^{- 9} - g (t) d t} = + 6

Janet Young

Beginner2021-12-15Added 32 answers

\int_{- 3}^{3} h (x) d x = 4

\int_{3}^{6} h (x) d x = - 12

\int_{- 3}^{6} h (t) d t

∵ \int_{- 3}^{6} h (t) d t = \int_{- 3}^{3} h (x) d x + \int_{3}^{6} h (x) d x

{\int_{- 3}^{6} h (t) d t = - 8 + \int_{6}^{- 3} h (t) d t = + 8}

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