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,