Exercise 3.3
Exercise 3.3 Exercise 3.3
MA111: Prepared by Dr.Archara Pacheenburawana 39 2. Use the form of the definition of the integral given in (4.7) to evaluate the integral. (a) (c) (e) ∫ 1 0 ∫ 5 1 ∫ 5 0 3. Prove that 4. Prove that 2xdx (2+3x−x 2 )dx (1+2x 3 )dx ∫ b a ∫ b a xdx = b2 −a 2 . 2 x 2 dx = b3 −a 3 . 3 (b) (d) (f) ∫ 5 −1 ∫ 2 0 ∫ 2 1 (1+3x)dx (2−x 2 )dx x 3 dx 5. Write the given sum or difference as a single integral in the form ∫ b f(x)dx. a 6. If 7. If (a) (c) (e) 1. (a) ∫ 8 2 ∫ 1 ∫ 2 0 ∫ 2 0 ∫ 3 f(x)dx+ f(x)dx+ f(x)dx+ ∫ 3 2 ∫ 1 2 ∫ 6 1 3 f(x)dx = 1.8 and f(x)dx = 3,, ∫ 4 f(x)dx f(x)dx f(x)dx+ ∫ 8 5 (b) (d) ∫ 3 0 ∫ 2 f(x)dx− f(x)dx+ ∫ 3 2 ∫ 3 −1 2 ∫ 12 6 f(x)dx f(x)dx = 3.2, find f(x)dx = −7, and ∫ 4 ∫ 5 2 f(x)dx. f(x)dx = 2, find f(x)dx f(x)dx 0 0 3 1 ∫ π 0 xsinxdx (b) Answer to Exercise 4.3 ∫ 5 1 e x ∫ 1 1+x dx (c) 2. (a) 1 (b) 42 (c) 2 (d) 4 3 (e) 317.5 (f) 3.75 5. (a) ∫ 3 f(x)dx (b) ∫ 2 f(x)dx (c) ∫ 1 (2x 2 −5x)dx (d) 0 1 f(x)dx (d) ∫ 3 ∫ 4 f(x)dx. √ xdx 0 0 0 −1 1 6. −1.4 7. −12 ∫ 3 f(x)dx (e) ∫ 12 f(x)dx
MA111: Prepared by Dr.Archara Pacheenburawana 40 Exercise 4.4 1. Evaluate the definite integrals using the First Fundamental Theorem of Calculus. (a) (c) (e) (g) (i) (k) (m) (o) (q) (s) (t) ∫ 3 −1 ∫ 4 0 ∫ 2 1 ∫ 3 3 ∫ π π/2 ∫ 9 1 ∫ 9 8 ∫ π/2 π/6 ∫ 2 0 ∫ 3 −2 ∫ 2 0 x 5 dx (b) ∫ 2 0 (2x−3)dx ∫ √ 4 xdx (d) ( √ x+3x)dx ∫ 3 1 x dx (f) (x √ x+x −1/2 )dx 4 0 √ ∫ π/2 x5 +2dx (h) 2sinxdx secxtanxdx (j) 0 0 ∫ π π/2 (2sinx−cosx)dx ∫ 1 1 2x dx (l) (e x −e −x )dx 2 x dx (n) ( x+ 2 ) sin 2 dx x (p) |2x−3|dx (r) 0 ∫ 3 0 ∫ √ 3 1 ∫ 3π/4 0 (3e x −x 2 )dx 6 1+x 2 dx |cosx|dx { −x, x ≥ 0 f(x)dx, where f(x) = x 2 , x < 0 { x f(x)dx, where f(x) = 4 , 0 ≤ x < 1 x 5 , 1 ≤ x ≤ 2 2. Use the Second Fundamental Theorem of Calculus to find the derivative. ∫ x ∫ x √ (a) f(x) = (t 2 −3t+2)dt (b) g(x) = 1−2tdt (c) g(y) = (e) f(x) = (g) h(x) = (i) y = ∫ 1 0 ∫ y 1−3x 2 ∫ x 2 0 ∫ 1/x 2 t 2 sintdt (d) F(x) = (e −t2 +1)dt (f) f(x) = arctantdt (h) y = 0 ∫ 2 x ∫ −1 x ∫ √ x u 3 ∫ x 3 1+u du (j) y = 2 3 cos(t 2 )dt cost t ln(t 2 +1)dt dt √ x √ tsintdt
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MA111: Prepared by Dr.Archara Pacheenburawana 40<br />
<strong>Exercise</strong> 4.4<br />
1. Evaluate the definite integrals using the First Fundamental Theorem of Calculus.<br />
(a)<br />
(c)<br />
(e)<br />
(g)<br />
(i)<br />
(k)<br />
(m)<br />
(o)<br />
(q)<br />
(s)<br />
(t)<br />
∫ 3<br />
−1<br />
∫ 4<br />
0<br />
∫ 2<br />
1<br />
∫ 3<br />
3<br />
∫ π<br />
π/2<br />
∫ 9<br />
1<br />
∫ 9<br />
8<br />
∫ π/2<br />
π/6<br />
∫ 2<br />
0<br />
∫ 3<br />
−2<br />
∫ 2<br />
0<br />
x 5 dx<br />
(b)<br />
∫ 2<br />
0<br />
(2x−3)dx<br />
∫<br />
√ 4<br />
xdx (d) ( √ x+3x)dx<br />
∫<br />
3<br />
1<br />
x dx (f) (x √ x+x −1/2 )dx<br />
4 0<br />
√<br />
∫ π/2<br />
x5 +2dx (h) 2sinxdx<br />
secxtanxdx<br />
(j)<br />
0<br />
0<br />
∫ π<br />
π/2<br />
(2sinx−cosx)dx<br />
∫<br />
1<br />
1<br />
2x dx (l) (e x −e −x )dx<br />
2 x dx (n)<br />
(<br />
x+ 2 )<br />
sin 2 dx<br />
x<br />
(p)<br />
|2x−3|dx<br />
(r)<br />
0<br />
∫ 3<br />
0<br />
∫ √ 3<br />
1<br />
∫ 3π/4<br />
0<br />
(3e x −x 2 )dx<br />
6<br />
1+x 2 dx<br />
|cosx|dx<br />
{ −x, x ≥ 0<br />
f(x)dx, where f(x) =<br />
x 2 , x < 0<br />
{ x<br />
f(x)dx, where f(x) =<br />
4 , 0 ≤ x < 1<br />
x 5 , 1 ≤ x ≤ 2<br />
2. Use the Second Fundamental Theorem of Calculus to find the derivative.<br />
∫ x<br />
∫ x √<br />
(a) f(x) = (t 2 −3t+2)dt (b) g(x) = 1−2tdt<br />
(c) g(y) =<br />
(e) f(x) =<br />
(g) h(x) =<br />
(i) y =<br />
∫ 1<br />
0<br />
∫ y<br />
1−3x<br />
2<br />
∫ x 2<br />
0<br />
∫ 1/x<br />
2<br />
t 2 sintdt (d) F(x) =<br />
(e −t2 +1)dt (f) f(x) =<br />
arctantdt (h) y =<br />
0<br />
∫ 2<br />
x<br />
∫ −1<br />
x<br />
∫ √ x<br />
u 3<br />
∫ x 3<br />
1+u du (j) y = 2<br />
3<br />
cos(t 2 )dt<br />
cost<br />
t<br />
ln(t 2 +1)dt<br />
dt<br />
√ x<br />
√<br />
tsintdt