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Riemann Sums / Definite Integral Worksheet

Riemann Sums / Definite Integral Worksheet
  • Page 1
 1.  
Evaluate the definite integral 02(5 + x) dx.
a.
1
b.
12
c.
- 12
d.
2


Solution:

f(x) = 5 + x
[Write the function.]

Since f is continuous on the interval [0, 2], f is integrable on [0, 2].

Δ xi = Δx = b-an = 2-0n = 2n
[Find the width of the ith subinterval.]

ci = a + ix) = 0 + i(2n) = 2in
[ci is the right end point of each subinterval.]

02(5 + x) dx = lim||Δ||0 i=1n f(cixi
[Write the definite integral as the limit of Riemann sum.]

= limn i=1n(5 + 2in)(2n)

= limn (2n) i=1n(5 + 2in)

= limn (2n) {5n + 2n[n(n+1)2]}
[Use i=1ni = n(n+1)2]

= 12
[Evaluate the limit.]

Therefore, 02(5 + x) dx = 12


Correct answer : (3)
 2.  
Evaluate the definite integral 03ex dx.
a.
e(e2 - 1)
b.
e
c.
e3 - 1
d.
1 - e3


Solution:

f(x) = ex
[Write the function.]

Since f is continuous on the interval [0, 3], f is integrable on [0, 3].

Δ xi = Δx = b-an = 3-0n = 3n
[Find the width of the ith subinterval.]

ci = a + ix) = 3in
[ci is the right end point of each subinterval.]

03ex dx = lim||Δ||0 i=1n f(cixi
[Write the definite integral as the limit of Riemann sum.]

= limn i=1ne3in (3n)

= limn (3n) [e3n+e2(3n)+....+en(3n)]

= limn(3n)[e(3n)[(e3n)n-1]e3n-1]

= lim1/n0e3n(e3-1)lim1/n0(e3n-13n)

= e3 - 1
[Use limx0 ex-1x = 1.]

Therefore, 03ex dx = e3 - 1


Correct answer : (4)
 3.  
Evaluate the definite integral 011 3 x2 dx.
a.
1 9
b.
1
c.
1 3
d.
2 3


Solution:

f(x) = 1 / 3 x2
[Write the function.]

Since f is continuous on the interval [0, 1], f is integrable on [0, 1].

Δ xi = Δx = b-an = 1-0n = 1n
[Find the width of the ith subinterval.]

ci = a + ix) = 0 + i(1n) = in
[ci is the right end point of each subinterval.]

011 / 3 x2 dx = lim||Δ||0 i=1n f(cixi
[Write the definite integral as the limit of Riemann sum.]

= limn i=1n1 / 3 (in)2(1n)

= limn (13n) i=1n(i2n2)

= limn (13n3) [n (n+1) (2n+1)6]
[Use i=1ni2 = n(n+1)(2n+1)6]

= 1 / 9
[Evaluate the limit.]

Therefore, 011 / 3 x2 dx = 1 / 9


Correct answer : (1)
 4.  
Evaluate the definite integral 01 (x2 + x) dx.
a.
5 6
b.
2
c.
5
d.
3


Solution:

f(x) = x2 + x
[Write the function.]

Since f is continuous on the interval [0, 1], f is integrable on [0, 1].

Δ xi = Δx = b-an = 1-0n = 1n
[Find the width of the ith subinterval.]

ci = a + ix) = in
[ci is the right end point of each subinterval.]

01(x2 + x) dx = lim||Δ||0 i=1n f(cixi
[Write the definite integral as the limit of Riemann sum.]

= limn i=1n[(in)2 + in)](1n)

= limn (1n) i=1n(i2n2+in)

= limn (1n) [1n2(n (n+1) (2n+1)6)+1n(n (n+1)2)]
[Use i=1ni2 = n(n+1)(2n+1)6 and i=1ni = n (n+1)2.]

= 1 / 3 + 1 / 2 = 5 / 6
[Evaluate the limit.]

Therefore, 01(x2 + x) dx = 5 / 6


Correct answer : (1)
 5.  
Evaluate the definite integral 02(2x + 1) dx.
a.
2
b.
5
c.
- 5
d.
6


Solution:

f(x) = 2x + 1
[Write the function.]

Since f is continuous on [0, 2], f is integrable on the interval [0, 2].

Δxi = Δx = b-an = 2-0n = 2n
[Find the width of ith sub interval.]

ci = a + ix) = 0 + i(2n) = 2in
[ ci is the right end point of each sub interval.]

02(2x + 1) dx = lim||Δ||0 f(ci) Δxi
[Write the definite integral as the limit of Riemann sum.]

= limn i=1n [2(2in) + 1](2n)

= limn (2n) i=1n (4in + 1)

= limn (2n) [4n(n(n+1)2) + n]
[Use i=1n i = n(n+1)2.]

= 6
[Evaluate the limit.]

Therefore, 02 (2x + 1) dx = 6


Correct answer : (5)
 6.  
Evaluate the definite integral 12 (x2 + x + 1) dx
a.
11
b.
29 6
c.
4
d.
35 3


Solution:

f(x) = x2 + x + 1
[Write the function.]

Since f is continuous on [1, 2], f is integrable on the interval [1, 2].

Δxi = Δx = b-an = 1n
[Find the width of ith sub interval.]

ci = a + ix) = 1 + i(1n) = 1 + in
[ci is the right end point of each sub interval.]

12 (x2 + x + 1) dx = lim||Δ||0 i=1n f(ci) Δxi
[Write the definite integral as the limit of Riemann sum.]

= limn i=1n [(1 + in)2 + (1 + in) + 1] (1n)

= limn (1n) i=1n [3 + 3in+i2n2]

= limn (1n) [3n + 3nn(n+1)2+1n2(n(n+1)(2n+1)6)]

= 3 + 3 / 2 + 1 / 3 = 29 / 6
[Evaluate the limits.]

Therefore, 12 (x2 + x + 1) dx = 29 / 6


Correct answer : (2)
 7.  
Evaluate the definite integral 01 (x2 - 2x - 3) dx
a.
2
b.
1
c.
- 1
d.
- 11 3


Solution:

f(x) = x2 - 2x - 3
[Write the function.]

Since f is continuous on [0, 1], f is integrable on the interval [0, 1].

Δxi = Δx = b-an = 1-0n = 1n
[Find the width of the ith sub interval.]

ci = a + ix) = 0 + i(1n) = in
[ci is the right end point of each sub interval.]

01 (x2 - 2x - 3) dx = lim||Δ||0 f(ci) Δxi

= limn i=1n [(in)2 - 2(in) - 3] (1n)

= limn (1n) i=1n [i2n2-2in - 3]

= limn (1n) [1n2(n(n+1)(2n+1)6)-2n(n(n+1)2) - 3n]

= 1 / 3 - 1 - 3 = - 11 / 3
[Evaluate the limit.]

Therefore, 01 (x2 - 2x - 3) dx = - 11 / 3


Correct answer : (4)
 8.  
Evaluate the definite integral -11 (x3 - 2x) dx
a.
1
b.
- 1
c.
- 2
d.
2


Solution:

f(x) = x3 - 2x
[Write the function.]

Since f is continuous over [- 1, 1], f is integrable on the interval [- 1, 1].

Δxi = Δx = b-an = 1-(-1)n = 2n
[Find the width of ith sub interval.]

ci = a + ix) = - 1 + i(2n) = - 1 + 2in
[ci is the right end point of each sub interval.]

-11 (x3 - 2x) dx = lim||Δ||0 i=1n f(ci) Δxi

= limn i=1n [(- 1 + 2in)3 - 2(- 1 + 2in)] (2n)

= limn (2n) i=1n [1 + 2in-12i2n2+8i3n3]

= limn (2n) [n + 2n(n(n+1)2)-12n2(n(n+1)(2n+1)6)+8n3(n2(n+1)24)]

= 2(1 + 1 - 4 + 2) = 0
[Evaluate the limit.]

Therefore, -11 (x3 - 2x) dx = 0


Correct answer : (3)
 9.  
Evaluate the definite integral -10 (x3 + 1) dx
a.
- 3
b.
3 4
c.
1
d.
- 1


Solution:

f(x) = (x3 + 1)
[Write the function.]

Since f is continuous on [- 1, 0], f is integrable on the interval [- 1, 0].

Δxi = Δx = b-an = 0-(-1)n = 1n
[Find the width of ith sub interval.]

ci = a + (Δx) = - 1 + i(1n) = - 1 + in
[ci is the right end point of each sub interval.]

-10 (x3 + 1) dx = lim||Δ||0 i=1n f(ci) Δxi

= limn i=1n [(- 1 + in)3 + 1] (1n)

= limn (1n) i=1n [i3n3+3in-3i2n2]

= limn (1n) [1n3(n2(n+1)24)+3n(n(n+1)2)-3n2(n(n+1)(2n+1)6)]

= 1 / 4 - 1 + 3 / 2 = 3 / 4
[Evaluate the limit.]

Therefore, -10 (x3 + 1) dx = 3 / 4


Correct answer : (2)
 10.  
Evaluate the definite integral 04 (1 - 3x + x2 - 2x3) dx
a.
- 211
b.
- 124
c.
- 380 3
d.
- 1876


Solution:

f(x) = (1 - 3x + x2 - 2x3)
[Write the function.]

Since f is continuous on [0, 4], f is integrable on the interval [0, 4].

Δxi = Δx = b-an = 4-0n = 4n
[Find the width of the ith sub interval.]

ci = a + ix) = 4in
[ci is the right end point of each sub interval.]

04 (1 - 3x + x2 - 2x3) dx = lim||Δ||0 i=1n f(ci) Δxi

= limn i=1n [1 - 3(4in) + (4in)2 - 2(4in)3] (4n)

= limn (4n) [n - 12n(n(n+1)2)+16n2(n(n+1)(2n+1)6)-128n3(n2(n+1)24)]

= 4 [1 - 6 + 16 / 3 - 32] = - 380 / 3
[Evaluate the limit.]

Therefore, 04 (1 - 3x + x2 - 2x3) dx = - 380 / 3


Correct answer : (3)

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