FIND THE VALUES OF C THAT SATISFY THE MEAN VALUE THEOREM FOR INTEGRALS

If f is continuous on [a, b], then at some point c in [a, b],

mean-value-theorem-of-integral

The graphical rectangular interpretation of the Mean value theorem for Definite Integrals is that:

If f is continuous on [a, b], then at some point c in [a, b] there is a rectangle with height f(c), and length b – a, such as the area of the rectangle equals the area under the curve f(x) on the interval [a, b]

For each problem, find the values of c that satisfy the Mean Value Theorem for Integrals.

Problem 1 :

f(x) = x + 2; [-3, 2]

Solution :

Finding average value of the function :

Here a = -3, b = 2

fave = 1b - a baf(x)dxfave = 12 + 3 2-3(x + 2)dx= 15 x22+ 2x2-3= 15(2)22+ 2(2) - (-3)22 + 2(-3) = 1542 + 4 - 92 + 6= 1542 + 82 -92 + 122= 15152= 32= 1.5

Finding the value of c that lies in the given interval :

f(x) = x + 2

f(c) = c + 2

1.5 = c + 2

c = 1.5 - 2

c =  -0.5

So, the value of c is -0.5, which lies in the interval.

Problem 2 :

f(x) = x22 - 3x + 52; [3, 5]

Solution :

Finding average value of the function :

Here a = 3, b = 5

fave = 1b - a baf(x)dxfave = 15 - 3 53x22 - 3x + 52dx= 12 x36- 3x22+ 5x253= 12(5)36 - 3(5)22+ 5(5)2 - (3)36 - 3(3)22+ 5(3)2 = 121256 - 752 + 252 - 276 + 272 - 152= 121256 - 502 -276 + 122= 121256 - 25 - 276 - 6= 12986 - 19= 12493 - 19= 1249 - 573= 12 × -83= -43

Equating the value derived from mean value theorem for integrals to f(x), we get

c22 - 3c + 52 = -43 c22 - 3c + 52 + 43 = 03c26 - 18c6 + 15 6 + 86 = 03c2 - 18c + 15 + 8 = 03c2 - 18c + 23 = 0x = -b ± b2 - 4ac2aa = 3, b = -18, c = 23c = -(-18) ± (-18)2 - 4(3)(23)2(3)= 18 ± 324 -276 6= 18 ± 48 6 = 18 ± 16(3) 6= 18 ± 43 6= 9 ± 233= 93 ± 233c =93 + 233, c =93 - 233 (not in the given interval)c =93 + 233

So, the value of c is (9 + 2√3)/3, which lies in the interval.

Problem 3 :

f(x) = 5x2; [2, 3]

Solution :

Finding average value of the function :

Here a = 2, b = 3

fave = 1b - a baf(x)dxfave = 13 - 2 325x2dx= 11325x2dx= 325x2dx= 5xx232= 5x32= 53 - 52= 10 - 156= -56

Equating the value derived from mean value theorem for integrals to f(x), we get

5c2 = -56-5c2= 30 c2 = -305c2 = -6c = 6

So, the value of c is √6, which lies in the interval.

Problem 4 :

f(x) = x12; [0, 3]

Solution :

Finding average value of the function :

Here a = 0, b = 3

fave = 1b - a baf(x)dxfave = 13 - 0 30x12dx= 1330x12dx= 13x12 + 112 +130= 13x323230= 1333232 - 03232 = 1333232= 13332 × 23= 69= 23

Equating the value derived from mean value theorem for integrals to f(x), we get

f(c) = c1223 = c12c = 43

So, the value of c is 4/3, which lies in the interval.

Problem 5 :

f(x) = -5(x + 1)2; [1, 3]

Solution :

Finding average value of the function :

Here a = 1, b = 3

fave = 1b - a baf(x)dxfave = 13 - 1 31-5(x + 1)2dx= 1231-5(x + 1)2dx= -52311(x + 1)2 dx= -52(x + 1)-2 + 1-2 + 131= -52(x + 1)-1-131= -521x + 131= -5213 + 1 - 11 + 1= -5214 - 12= -521 - 24= -52-14= 58

Equating the value derived from mean value theorem for integrals to f(x), we get

f(c) = 58f(x) = -5(x + 1)258 = -5(c + 1)25(c + 1)2 = -405c2 + 2c + 1 = -405c2 + 10c + 5 + 40 = 05c2 + 10c + 45 = 05c2 + 2c + 9 = 0c2 + 2c + 9 = 0a = 1, b = 2, c = 9c = -(2) ± (2)2 - 4(1)(9)2(1)= -2 ± 4 -362= -2 ± 32 2 = -2 ± 16(2) 2= -2 ± 42 2= -21 ± 222= -1 ± 22c = -1 + 22, -1 - 22c = -1 + 22

So, the value of c is -1 + 2√2 , which lies in the interval.

Problem 6 :

f(x) = (x - 1)12; [1, 4]

Solution :

Finding average value of the function :

Here a = 1, b = 4

fave = 1b - a baf(x)dxfave = 14 - 1 41(x - 1)12dx= 13 (x - 1)12 + 112 + 141= 13 (x - 1)32 32 41= 13 (4 - 1)32 32 = 13 332 32 = 13 332 ×23 = 13 63= 23

Equating the value derived from mean value theorem for integrals to f(x), we get

f(c) = (c - 1)12f(c) = 2323 = (c - 1)12c - 1 = 23 × 21c - 1 = 43c = 43 + 1c = 73

So, the value of c is 7/3 , which lies in the interval.

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