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The Fundamental Theorem of Calculus

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1. Use Part 1 of the FTC to find the derivative of the function
a) y=∫ from 1 to 2x    (sin(t))/(t) dt
b) g(x)= ∫ sinx to 0  sqrt(t) dt
c) f(x)= ∫ sqrt(x) to x^2  ((e^(-t)+e^(t))/(t)) dt
 
2. Evaluate the definite integral. Give exact solutions.
a) ∫ 1 to 9   (x−(2/sqrt(x))^2
b) ∫ 0 to (pi/3)  secx(secx-tanx) dx
c) ∫ (pi/6) to (pi/3)  (cos2x)/(cos^(2)x) dx

 

asked Apr 19, 2015 in CALCULUS by anonymous

6 Answers

0 votes

1. (a)

Step 1 :

The fundamental theorem of calculus, part 1 :

If  is continuous on then the function image is defined by image

is continuous on and differentiable on , and .

Step 2 :

The equation is

Compare with

The function .

Thus, from the fundamental theorem of calculus, part 1 image

The derivative of the function is,

image

image.

The derivative of the function image.

Solution :

The derivative of the function image.

answered Apr 20, 2015 by Sammi Mentor
edited Apr 20, 2015 by Sammi
0 votes

1. (b)

Step 1 :

The fundamental theorem of calculus, part 1 :

If  is continuous on then the function image is defined by image

is continuous on and differentiable on , and .

Step 2 :

The equation is image.

Apply definition of special definite integrals: .

image.

image

image.

image

Apply derivative on each side.

image

image

image

Substitute image and image.

image.

image.

The derivative of the function image.

Solution :

The derivative of the function image.

answered Apr 20, 2015 by Sammi Mentor
edited Apr 20, 2015 by Sammi
0 votes

Step 1:

2. (a)

The integral is .

image

image.

Solution :

image.

answered Apr 20, 2015 by joseph Apprentice
0 votes

Step 1:

2. (b)

The integral is image.

image

image.

Solution :

image.

answered Apr 20, 2015 by joseph Apprentice
0 votes

Step 1:

2. (c)

The integral is image.

image

image.

Solution :

image.

answered Apr 20, 2015 by joseph Apprentice
reshown Apr 20, 2015 by casacop
0 votes

1. (c)

Step 1 :

The fundamental theorem of calculus, part 1 :

If  is continuous on then the function image is defined by image

is continuous on and differentiable on , and .

Step 2 :

The equation is image.

Apply Additive interval property: image.

image

Apply definition of special definite integrals: .

image

Apply derivative on each side.

image

image

image

image

image

image.

The derivative of the function image.

Solution :

The derivative of the function image.

answered Apr 20, 2015 by Sammi Mentor

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