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Chapter 4: Q. 5 (page 364)

Find a function f that has the given derivative f and value f(c), if possible.

f'(x)=53x-2,f(1)=8

Short Answer

Expert verified

The function f is f(x)=5ln3x-2+c and the value of (c) is 8.

Step by step solution

01

Step 1. Given Information.

The derivative:

f'(x)=53x-2,f(1)=8

02

Step 2. Find the function by integration.

Integrating both sides with respect to x,

鈭�f'(x)dx=鈭�53x-2dx=5鈭�13x-2dx=5ln3x-2+c

where c is a constant.

03

Step 3. Find the value of c

Substitute the values in the function,

f(x)=5ln3x-2+c8=5ln3(1)-2+c8=5ln1+cc=8

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Most popular questions from this chapter

Determine which of the limit of sums in Exercises 47鈥�52 are infinite and which are finite. For each limit of sums that is finite, compute its value

limn鈫�鈭�鈭�k=1n1+kn2.1n

Prove Theorem 4.13(b): For any real numbers a and b, we have鈭�abxdx=12b2-a2. Use the proof of Theorem 4.13(a) as a guide.

What is the difference between an antiderivative of a function and the indefinite integral of a function?

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

鈭�鈭�32鈥�x2+52xdx

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|鈭�abf(x)dx|.

(b) True or False: The area of the region between f(x) = x 鈭� 4 and g(x) = -x2on the interval [鈭�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 鈭�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].
See all solutions

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