crms calculus 2010 may 12, 2010

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1 Height of the Riemann rectangle which is yvalue of f is negative. The product of positive(dx) and negative(f(x)) value is negative. Therefore, the value of the first integral is negative. Height of the Riemann rectangle which is yvalue of f is positive. The product of positive(dx) and positive(f(x)) value is positive. Therefore, the value of the first integral is positive. The integral∫f(x)dx is: positive if f(x) is positive for all xvalue in the interval [a,b] negative if f(x) is negative for all xvalue in the interval [a,b] provided a < b since the rectangle is the base times the height and the height of the rectangles are positive and because the bvalue is smaller than the avalue the change in x is negative and a negative base times a positive height equals a negative product (area)

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Properties of definite integrals

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Page 1: CRMS Calculus 2010 May 12, 2010

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Height of the Riemann rectangle which is y­value of  f is negative. The product of positive(dx) and negative(f(x)) value is negative. Therefore, the value of the first integral is negative. 

Height of the Riemann rectangle which is y­value of  f is positive. The product of positive(dx) and positive(f(x)) value is positive. Therefore, the value of the first integral is positive. 

The integral∫f(x)dx is:positive if f(x) is positive for all x­value in the interval [a,b]negative if f(x) is negative for all x­value in the interval [a,b]provided a < b

since the rectangle is the base times the height and the height of the rectangles are positive and because the b­value is smaller than the a­value the change in x is negative and a negative base times a positive height equals a negative product (area)

Page 2: CRMS Calculus 2010 May 12, 2010

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Symmetric over the y­axis

The first integral is twice the second due the fact that it is an even function (meaning the two sides are equal) and our regions are equidistant from the y­axis. The first contains both sides of the function while the second is only half, or one side of the function.

Page 3: CRMS Calculus 2010 May 12, 2010

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Symmetrical over the origin

The answer is zero because the area ­4 to 0 is negative and the area 0 to 4 is positive but they are the same value because it is an odd function so they subtract to zero. 

If f(x) is an even function then: If f(x) is an odd function then:

Page 4: CRMS Calculus 2010 May 12, 2010

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see graph

Each y­value of h(x) is doubled to build the graph of 2h(x).

Enter h(x) into Y1.

2Y1(6) ­ 2Y1(1) = 54

see graph

54 is double 27.

5(27) = 135

to the constant times the integral of the function

The integral of a constant times a function 

is equal

Page 5: CRMS Calculus 2010 May 12, 2010

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