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Use the Fundamental Theorem of Calculus to give alternative proofs of the integration facts shown in Exercises 72–76. You may assume that all functions here are integrable

∫abxdx=12(b2-a2)

Short Answer

Expert verified

Proved that ∫abxdx=12(b2-a2)

Step by step solution

01

Step 1. Given information

The given integral∫abxdx=12(b2-a2)

02

Step 2. Prove that ∫abx dx=12(b2-a2) using the fundamental theorem of calculas

∫abxdx=x22ab=12b2-12a2=12b2-a2=12(b2-a2)

Therefore,it is proved that∫abxdx=12(b2-a2)

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Most popular questions from this chapter

Show by exhibiting a counterexample that, in general, ∫f(x)g(x)dx≠∫f(x)dx∫g(x)dx. In other words, find two functions f and g so that the integral of their product is not equal to the product of their integrals.

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer.

∫−32 1x+5dx

If ∫-23f(x)dx=4,∫-26f(x)dx=9,∫-23g(x)dx=2and ∫36g(x)dx=3,then find the values of each definite integral in Exercises 29-40. If there is not enough information, explain why.

∫36fxdx.

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The absolute area between the graph of f and the x-axis on [a, b] is equal to|∫abf(x)dx|.

(b) True or False: The area of the region between f(x) = x − 4 and g(x) = -x2on the interval [−3, 3] is negative.

(c) True or False: The signed area between the graph of f on [a, b] is always less than or equal to the absolute area on the same interval.

(d) True or False: The area between any two graphs f and g on an interval [a, b] is given by ∫ab(f(x)-g(x))dx.

(e) True or False: The average value of the function f(x) = x2-3 on [2, 6] is

f(6)+f(2)2= 33+12= 17.

(f) True or False: The average value of the function f(x) = x2-3on [2, 6] is f(6)-f(2)4= 33-14= 8.

(g) True or False: The average value of f on [1, 5] is equal to the average of the average value of f on [1, 2] and the average value of f on [2, 5].

(h) True or False: The average value of f on [1, 5] is equal to the average of the average value of f on [1, 3] and the average value of f on [3, 5].

Write each expression in Exercises 41–43 in one sigma notation (with some extra terms added to or subtracted from the sum, as necessary).

3∑k=225k2+2∑kk=224-∑k=0251

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