46-antiderivatives

  • November 2019
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The derivative of xn is nxn−1. The derivative of xn+1 is (n + 1)xn. The derivative of

xn+1 n+1

is xn, if n 6= −1.

An antiderivative of xn is An antiderivative of x4 is

xn+1 n+1 x5 5

, if n 6= −1.

.

Another antiderivative of x4 is

x5 5

Yet another antiderivative of x4 is

+ 13. x5 5

− 1.

If C is any constant, the derivative of If C is any constant,

x5 5

x5 5

+ C is x4.

+ C is an antiderivative of x4.

Indeed, any antiderivative of x4 is of the form

x5 5

+ C.

An Application of the Mean Value Theorem: If f 0(x) = g 0(x) for all x in an interval, then f (x) = g(x) + C for a constant C. In general, we also have Any antiderivative of xn is of the form

xn+1 n+1

+ C, if n 6= −1.

Any antiderivative of xn is of the form

xn+1 n+1

Any antiderivative of x2 − 1 is of the form

x3

+ C, if n 6= −1. − x + C.

3 To find the antiderivative f (x) of x − 1 satisfying f (1) = 3 : 2

f (x) = f (1) = 3 = 11 3

x3

3 13 3 1 3

−x+C −1+C

−1+C

= C

f (x) =

x3 3

−x−

11 3

xn+1

+ C, if n 6= −1. Any antiderivative of xn is of the form n+1 √ √ 1 1 3 2 3 Any antiderivative of x − x = x − x 2 is of the form

x

1 3

1 +1 3

+1



x

1 2

1 +1 2

+1

+C = √ 3

x

4 3

4 3



x

3 2

3 2

+ C.

√ The only antiderivative f (x) of x− 2 x satisfying f (0) = 3 : 2 3 3 4 3 f (x) = x − x 2 + C. 4 3 2 3 3 4 f (0) = 0 3 − 0 2 + C. 4 3 3 = C 2 3 3 4 3 f (x) = x − x 2 + 3 4 3

Any piecewise-antiderivative of f (x) =

   1      2     −1

for x < 1, for 1 < x < 3, for 3 < x < 4,

 x − 4 for 4 < x < 5,       6 − x for 5 < x < 6,     0 for 6 < x.    x + C1 for x < 1,      for 1 < x < 2x + C2     −x + C for 3 < x < 3 may be written as G(x) = 1 2  x − 4x + C4 for 4 < x <   2 with G(1), G(3), G(4),   1 2   x + 6x + C5 for 5 < x < −  G(5) and G(6) 2    C6 taking any values. for 6 < x.

3, 4, 5, 6,

   x + C1      2x + C2     −x + C 3 G(x) = 1 2  x − 4x + C4   2   1 2   x + 6x + C5 −  2    C6

for x < 1, for 1 < x < 3, for 3 < x < 4, for 4 < x < 5, for 5 < x < 6, for 6 < x.

To have G continuous at x = 1 we need x + C1 = 2x + C2. To have G continuous at x = 3 we need 2x + C2 = −x + C3. Similarly, at x = 4 we need −x + C3 = 12 x2 − 4x + C4. Similarly, at x = 5 we need 12 x2 − 4x + C4 = − 12 x2 + 6x + C5. To have G continuous at x = 6 we need − 12 x2 +6x+C5 = C6.

For G to be continuous we need 2x + C2 = x + C1 2 + C2 = 1 + C1 C2 = C1 − 1

at x = 1 :

at x = 3 :

at x = 4 :

−x + C3 = 2x + C2 −3 + C3 = 6 + C2 C3 = C2 + 9 = C 1 + 8 1 2 1 2

x2 − 4x + C4 = −x + C3 16 − 16 + C4 = −4 + C3 C4 = C3 + 4 = C1 + 12

For G to be continuous we need 1 2 1 2 at x = 5 : − x + 6x + C5 = x − 4x + C4 2 2 25 25 − 20 + C4 − + 30 + C5 = 2 2 C5 = C4 − 25 = C1 − 13 at x = 6 :

1

C6 = − x2 + 6x + C5 2 C6 = −18 + 36 + C5 C6 = C5 + 18 = C1 + 5

Recall from the Extended Mean Value Theorem: If two continuous piecewise-differentiable functions F and G satisfy F 0(x) = G0(x) for all but a finite number of x’s in an interval, then F (x) = G(x) + C for all x on that interval.

Any continuous piecewise-antiderivative G(x) of    1 for x < 1,      2 for 1 < x < 3,     −1 for 3 < x < 4, f (x) =  x − 4 for 4 < x < 5,       6 − x for 5 < x < 6,     0 for 6 < x.    x + C1 for x ≤ 1,      for 1 ≤ x ≤ 2x + C1 − 1     −x + C + 8 for 3 ≤ x ≤ 1 equals 1 2  x − 4x + C1 + 12 for 4 ≤ x ≤   2   1 2   x + 6x + C1 − 13 for 5 ≤ x ≤ −  2    C1 + 5 for 6 ≤ x.

3, 4, 5, 6,

Any continuous piecewise-antiderivative G(x) of    1 for x < 1,      2 for 1 < x < 3,     −1 for 3 < x < 4, f (x) =  x − 4 for 4 < x < 5,       6 − x for 5 < x < 6,     0 for 6 < x.    x for x ≤ 1,      2x − 1 for 1 ≤ x ≤     −x + 8 for 3 ≤ x ≤ equals C1 + 1 2  x − 4x + 12 for 4 ≤ x ≤   2   1 2   x + 6x − 13 for 5 ≤ x ≤ −  2    5 for 6 ≤ x.

3, 4, 5, 6,

The only continuous piecewise-antiderivative G(x) of    1 for x < 1,      2 for 1 < x < 3,     −1 for 3 < x < 4, which has G(0) = 0, f (x) =  x − 4 for 4 < x < 5,       6 − x for 5 < x < 6,     0 for 6 < x,    x for x ≤ 1,      2x − 1 for 1 ≤ x ≤ 3,     −x + 8 for 3 ≤ x ≤ 4, equals 1 (with C1 = 0.) 2  x − 4x + 12 for 4 ≤ x ≤ 5,   2   1 2   x + 6x − 13 for 5 ≤ x ≤ 6, −  2    5 for 6 ≤ x,

The only continuous piecewise-antiderivative F (x) of    1 for x < 1,      2 for 1 < x < 3,     −1 for 3 < x < 4, which has F (2) = 0, f (x) =  x − 4 for 4 < x < 5,       6 − x for 5 < x < 6,     0 for 6 < x.    x for x ≤ 1,      2x − 1 for 1 ≤ x ≤ 3,     −x + 8 for 3 ≤ x ≤ 4, equals − 3 + 1 = G(x) − 3. 2  x − 4x + 12 for 4 ≤ x ≤ 5,   2   1 2   x + 6x − 13 for 5 ≤ x ≤ 6, −  2    5 for 6 ≤ x,