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Find the value of the definite integral ∫_0^2x(x+1)dx.
Related Topics
Wize University Calculus 1 Textbook > Integrals
Properties of Definite Integrals
3 Activities
Find the value of the definite integral
∫
0
2
x
(
x
+
1
)
d
x
\displaystyle \int_0^2x(x+1)dx
∫
0
2
x
(
x
+
1
)
d
x
.
10/3
14/3
6
10
12
I don't know
Check Submission
More Properties of Definite Integrals Questions:
Properties of definite integrals
If
∫
0
8
f
(
x
)
d
x
=
5
,
∫
0
3
f
(
x
)
d
x
=
4
,
\int_0^8f(x)dx=5,\int_0^3f(x)dx=4,
∫
0
8
f
(
x
)
d
x
=
5
,
∫
0
3
f
(
x
)
d
x
=
4
,
and
∫
4
8
f
(
x
)
d
x
=
−
2
\int_4^8f(x)dx=-2
∫
4
8
f
(
x
)
d
x
=
−
2
, then what is the value of
∫
3
4
f
(
x
)
d
x
\int_3^4f(x)dx
∫
3
4
f
(
x
)
d
x
?
Evaluate the definite integral
∫
0
1
x
(
5
−
x
2
)
d
x
\int_0^1x(5-x^2)dx
∫
0
1
x
(
5
−
x
2
)
d
x
.
Evaluate the definite integral
∫
0
1
x
(
5
−
x
2
)
d
x
\int_0^1x(5-x^2)dx
∫
0
1
x
(
5
−
x
2
)
d
x
.
Evaluate the definite integral
∫
0
1
x
(
5
−
x
2
)
d
x
\int_0^1x(5-x^2)dx
∫
0
1
x
(
5
−
x
2
)
d
x
.
Find the definite integral
∫
0
2
x
(
x
+
1
)
d
x
\displaystyle\int_0^2x(x+1)dx
∫
0
2
x
(
x
+
1
)
d
x
.
Definite with Absolute Value
Evaluate the following integral:
∫
0
3
∣
x
2
−
x
−
2
∣
d
x
\displaystyle\int_0^3|x^2-x-2| \ dx
∫
0
3
∣
x
2
−
x
−
2∣
d
x
Definite Integral: Trigonometric Function and Absolute Value
Evaluate the definite integral:
∫
0
5
π
/
4
∣
sin
x
∣
d
x
\displaystyle\int_0^{5\pi/4} |\sin x| \ dx
∫
0
5
π
/4
∣
sin
x
∣
d
x
Practice: Definite with Fraction
Q:
\textbf{Q:}
Q:
Evaluate
∫
1
4
[
1
x
2
−
x
]
d
x
\displaystyle \int_1^4\left[\frac{1}{x^2}-x\right]dx
∫
1
4
[
x
2
1
−
x
]
d
x
Practice: Definite with Absolue Value
Q.
\textbf{Q.}
Q.
Compute the definite integral
∫
−
2
2
(
2
−
∣
x
∣
)
d
x
\displaystyle \int_{-2}^{2}(2-\lvert x\rvert)\text{d}x
∫
−
2
2
(
2
−
∣
x
∣)
d
x
Definite Integrals: Integration by Substitution
If
∫
−
2
7
g
(
x
)
d
x
=
13
\int_{-2}^7g\left(x\right)dx=13
∫
−
2
7
g
(
x
)
d
x
=
13
and
∫
5
7
2
g
(
x
)
=
10
\int_5^72g\left(x\right)=10
∫
5
7
2
g
(
x
)
=
10
, then
∫
−
1
5
2
g
(
2
x
)
d
x
=
\int_{-1}^{\frac{5}{2}}g\left(2x\right)dx=
∫
−
1
2
5
g
(
2
x
)
d
x
=
Properties of Definite Integrals
Suppose that
f
(
−
x
)
=
f
(
x
)
f\left(-x\right)=f\left(x\right)
f
(
−
x
)
=
f
(
x
)
and
g
(
−
x
)
=
−
g
(
x
)
g\left(-x\right)=-g\left(x\right)
g
(
−
x
)
=
−
g
(
x
)
. If
∫
−
1
0
2
f
(
x
)
=
5
\int_{-1}^02f\left(x\right)=5
∫
−
1
0
2
f
(
x
)
=
5
, the value of
∫
−
1
1
[
3
f
(
x
)
+
x
⋅
sin
x
⋅
g
(
x
)
]
d
x
=
\int_{-1}^1\left[3f\left(x\right)+x\cdot\sin x\cdot g\left(x\right)\right]dx=
∫
−
1
1
[
3
f
(
x
)
+
x
⋅
sin
x
⋅
g
(
x
)
]
d
x
=
Practice: Definite Integral
Evaluate
∫
−
2
0
∣
x
2
−
1
∣
d
x
\displaystyle \int_{-2}^0\left|x^2-1\right|dx
∫
−
2
0
x
2
−
1
d
x
Find
∫
0
π
∣
cos
x
∣
d
x
\int_0^{\pi}\left|\cos\ x\right|dx
∫
0
π
∣
cos
x
∣
d
x
Practice: Definite with Absolue Value
Q.
\textbf{Q.}
Q.
Compute the definite integral
∫
−
2
2
(
2
−
∣
x
∣
)
d
x
\displaystyle \int_{-2}^{2}(2-\lvert x\rvert)\text{d}x
∫
−
2
2
(
2
−
∣
x
∣)
d
x
Definite with Absolute Value
Evaluate the following integral:
∫
0
3
∣
x
2
−
x
−
2
∣
d
x
\displaystyle\int_0^3|x^2-x-2| \ dx
∫
0
3
∣
x
2
−
x
−
2∣
d
x
Definite Integral: Trigonometric Function and Absolute Value
Evaluate the definite integral:
∫
0
5
π
/
4
∣
sin
x
∣
d
x
\displaystyle\int_0^{5\pi/4} |\sin x| \ dx
∫
0
5
π
/4
∣
sin
x
∣
d
x
Practice: Definite with Fraction
Q:
\textbf{Q:}
Q:
Evaluate
∫
1
4
[
1
x
2
−
x
]
d
x
\displaystyle \int_1^4\left[\frac{1}{x^2}-x\right]dx
∫
1
4
[
x
2
1
−
x
]
d
x
Properties of definite integrals
If
∫
0
8
f
(
x
)
d
x
=
5
,
∫
0
3
f
(
x
)
d
x
=
4
,
\int_0^8f(x)dx=5,\int_0^3f(x)dx=4,
∫
0
8
f
(
x
)
d
x
=
5
,
∫
0
3
f
(
x
)
d
x
=
4
,
and
∫
4
8
f
(
x
)
d
x
=
−
2
\int_4^8f(x)dx=-2
∫
4
8
f
(
x
)
d
x
=
−
2
, then what is the value of
∫
3
4
f
(
x
)
d
x
\int_3^4f(x)dx
∫
3
4
f
(
x
)
d
x
?
Find the definite integral
∫
0
2
x
(
x
+
1
)
d
x
\displaystyle\int_0^2x(x+1)dx
∫
0
2
x
(
x
+
1
)
d
x
.
Evaluate the definite integral
∫
0
1
x
(
5
−
x
2
)
d
x
\int_0^1x(5-x^2)dx
∫
0
1
x
(
5
−
x
2
)
d
x
.
Antiderivatives: Trigonometric Functions
Evaluate
∫
π
3
π
∣
cos
x
∣
d
x
\int_{\frac{\pi}{3}}^{\pi}\left|\cos x\right|dx
∫
3
π
π
∣
cos
x
∣
d
x
.