Definite Integral
What is the area between the x-axis and the graph of the function f (x) = -x + 1 over the interval [–2, 3] ?
[–2, 3]
∫ (-x + 1) dx = (-1/2) x2 + x ;
area :
(-1/2) 32 + 3 - ((-1/2) (-2)2 + (-2)) = (-9/2) + 3 - ((-4/2) – 2)
= 5/2
What is the area between the x-axis and the graph of the function f (x) = (½) x2 + 4 over the interval [–3, 1] ?
[–3, 1]
Int[((1/2) x2 + 4) dx] = (1/6) x3 + 4x ;
area :
(1/6) 13 + 4 - ((1/6) (-3)3 + 4(-3)) = (1/6) + 4 - ((-27/6) – 12)
= 16 – (28/6) = 16 + (14/3)
= 62/3
For the interval [e, 4], ∫ (x2 + (6/x)) dx = ?
[e, 4]
(1/3) x3 - 6 lnx ;
(1/3) 43 + 6 ln4 - ((1/3) e3 + 6 lne)
= (1/3) 64 + 6 ln4 - ((1/3) e3 + 6) = (1/3) (64 - e3) + 6
= (1/3) (82 - e3)
For the interval [a, e], ∫ ((x/2) - (2/x)) dx = ?
[a, e]
(1/6) x3 - 2 lnx ;
(1/6) e3 - 2 lne - ((1/6) a3 - 2 lna)
= (1/6) (e3 - a3) + 2 – 2 lna
= (1/6) (e3 - a3) - 2 (1 – lna)
For the interval [2, 4], ∫ (x/2)1/2 dx = ?
[2, 4]
((21/2)/3) x3/2 ;
((21/2)/3) 43/2 - ((21/2)/3) 23/2 = ((21/2)/3) 23/2 (23/2 - 1)
= (4/3) (23/2 - 1)
For the interval [6, 7], ∫ 5x1/3 dx = ?
[6, 7]
5 (3/4) x4/3 ;
(15/4) (74/3 - 64/3)
For the interval [2, 4], ∫ (x - 2)2 dx = ?
[2, 4]
Change of variable : u = (x - 2) ; du = dx ;
∫ u2 du = (1/3) u3
= (1/3) (x - 2)3 ;
(1/3) ((4 - 2)3 - (2 - 2)3) = (1/3) 23
= 8/3
For the interval [a, b], ∫ (y - x)4 dx = ?
[a, b]
Change of variable : u = (y - x) ; du = -dx ;
∫ u4 du = (-1/5) u5
= (-1/5) (y - x)5 ;
(-1/5) ((b - x)5 - (a - x)5)
On the interval [-1, 1], ∫ f(x) dx = e/2 and ∫ g(x) dx = e/3. Then ∫ (4 f(x) – 5 g(x)) dx = ?
Linearity : 4(e/2) – 5(e/3) = 2e/12
= e/6
On the interval [-10, -5], ∫ f(x) dx = -5 and ∫ g(x) dx = -10. Then ∫ (51/2 f(x) – 31/2 g(x)) dx = ?
Linearity : 51/2 (-5) – 31/2 (-10)
= (-5) (51/2 – 2(3) 1/2 )
For the interval [-1, 1], ∫ f(x) dx] = -5 ; for the interval [-1, 0] ∫ f(x) dx = 5. Then for the interval [0, 1] ∫ f(x) dx = ?
Linearity : -5 = 5 + a ;
a = -10
For the interval [-5, -1], ∫ f(x) dx = -10 ; for the interval [-2, -1] ∫ f(x) dx = -6. Then For the interval [-5, -2], ∫ f(x) dx = ?
Linearity : -10 = a + (-6) ;
a = -4
y = 2x for the interval [-1, 1] ; y = 2 for the interval [1, 4]. Then for the interval [-1, 4], ∫ y(x) dx = ?
[-1, 1] : ∫ 2x dx = x2 ; a = 12 - (-1)2 = 0
[1, 4] : ∫ 2 dx = 2x ; b = 2 (4 – 1) = 6
[-1, 4] : c = a + b = 6
y = -3x for the interval [-2, -1] ; y = 3 for the interval [-1, 6]. Then for the interval [-2, 6], ∫ y(x) dx = ?
[-2, -1] : ∫ -3x dx = (-3/2) x2 ; a = (-3/2) ((-1)2 - (-2)2 ) = (-3/2) (1 – 4) = 9/2
[-1, 6] : ∫ 3 dx = 3x ; b = 3 (6 – )(-1)) = 21
[-2, 6] : c = a + b = 9/2 + 21 = 51/2
For x ≤ 1/2, what is the area between the curves y = 3x2 and y = 9x ?
intersection points : 3x2 = 9x ; 3x2 – 9x = 0 ; 3x (x – 3) = 0 ; x = {0, 1}
–> x : [0, ½]
∫ (3x2 – 9x) dx = x3 – (9/2) x2 ;
a : (1/2)3 – (9/2) (1/2)2 – 0) = (1/2)3 – 9(1/2)3
= (1/2)3 (1 – 9) = (1/2)3 (-8) = -1
|a| = 1
For x ≥ -10, what is the area between the curves y = -x, y = x + 10 ?
intersection points : -x = x + 10 ; x = -10 / 2 = -5
[-10 , -5]
∫ (-x – (x + 10)) dx = ∫ (-2x – 10) dx = -x2 – 10x ;
a = (-(-10)2 – 10 (-10) - (-(-5)2 – 10 (-5))) = -100 + 100 - (-25 + 50)) = -25
|a| = 25
What is the mean value of f (x) = 2x2 - 1 between x = 2 and x = 4 ?
[2, 4]
∫ (2x2 – 1) dx] (2, 4) = (2/3) x2 - x ;
(2/3) 42 – 4 - ((2/3) (22 – 2) = (2/3) (16 – 4) - 4 + 2
= (2/3) 12 + 6 = 8 - 2 = 6
m = (1/(4 - 2)) 6 = 3
What is the mean value of f (x) = -4x3 between x = 2 and x = 5 ?
[2, 5]
∫ -4x3 dx = -x4 ;
-54 – -(24) = -625 + 16 = -609
m = (1/(5 - 2)) (-609) = -609/3 = -203
What is the area under the graph of y = (1/x) + 1 on the interval e ≤ x ≤ e2 ?
[e, e2]
∫ ((1/x) + 1) dx = lnx + x ;
ln(e2) + e2 - (lne + e) = 2 + e2 - (1 + e)
= 1 – e + e2
What is the area under the graph of y = (e/x) - e on the interval e2 ≤ x ≤ e3 ?
What is the area under the graph of y = (e/x) - e on the interval e2 ≤ x ≤ e3?
[e2, e3]
∫ ((e/x) - e) dx = ∫ e ((1/x) - 1) dx] = e (lnx - x) ;
e ((ln(e3) - e3 ) - (ln(e2) - e2 ))
= e ((3 - e3 ) - (2 - e2 ))
= e (1 – e2 + e3 )