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Practice: Definite Integral
Related Topics
Wize University Calculus 1 Textbook > Integrals
Antiderivatives of Inverse Trig Functions
3 Activities
Wize University Calculus 1 Textbook > Integrals
Antiderivatives of Exponential Functions
4 Activities
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
π
4
−
8
\frac{\pi}{4}-8
4
π
−
8
π
12
+
7
\frac{\pi}{12}+7
12
π
+
7
π
2
+
8
ln
2
\frac{\pi}{2}+8\ln2
2
π
+
8
ln
2
1
18
2
+
7
\frac{1}{18^2}+7
1
8
2
1
+
7
None of the above
I don't know
Check Submission
More Antiderivatives of Inverse Trig Functions Questions:
Integration Medley
Practice Question
Evaluate
∫
x
4
+
1
x
2
+
1
d
x
{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}
∫
x
2
+
1
x
4
+
1
d
x
.
Evaluate the integral
∫
x
4
+
1
x
2
+
1
d
x
\displaystyle\int\frac{x^4+1}{x^2+1}dx
∫
x
2
+
1
x
4
+
1
d
x
.
Find the value of the definite integral
∫
π
/
6
π
/
3
(
sec
x
+
tan
x
)
sec
x
d
x
\displaystyle \int_{\pi/6 }^{\pi/3}(\sec x+\tan x)\sec x dx
∫
π
/6
π
/3
(
sec
x
+
tan
x
)
sec
x
d
x
.
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Practice: Definite Integral (~F2017 Final Q26)
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
Integration Medley
Practice Question
Evaluate
∫
x
4
+
1
x
2
+
1
d
x
{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}
∫
x
2
+
1
x
4
+
1
d
x
.
Integration Medley
Practice Question
Evaluate
∫
x
4
+
1
x
2
+
1
d
x
{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}
∫
x
2
+
1
x
4
+
1
d
x
.
Integration Medley
Practice Question
Evaluate
∫
x
4
+
1
x
2
+
1
d
x
{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}
∫
x
2
+
1
x
4
+
1
d
x
.
Evaluate
∫
−
1
0
d
x
1
−
x
2
\int_{-1}^0\frac{dx}{\sqrt{1-x^2}}
∫
−
1
0
1
−
x
2
d
x
Integration Medley
Practice Question
Evaluate
∫
x
4
+
1
x
2
+
1
d
x
{\displaystyle\int} \frac{x^4+1}{x^2+1}\de{x}
∫
x
2
+
1
x
4
+
1
d
x
.
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\displaystyle \int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
Antiderivatives: Inverse Trigonometric Functions
Evaluate
d
d
u
cos
−
1
u
sin
−
1
u
\displaystyle \frac{\text{d}}{\text{d}u} \frac{\cos^{-1}u}{\sin^{-1}u}
d
u
d
sin
−
1
u
cos
−
1
u
. It is not necessary to simplify.
Evaluate the integral
∫
x
4
+
1
x
2
+
1
d
x
\displaystyle\int\frac{x^4+1}{x^2+1}dx
∫
x
2
+
1
x
4
+
1
d
x
.
Find the value of the definite integral
∫
π
/
6
π
/
3
(
sec
x
+
tan
x
)
sec
x
d
x
\displaystyle \int_{\pi/6 }^{\pi/3}(\sec x+\tan x)\sec x dx
∫
π
/6
π
/3
(
sec
x
+
tan
x
)
sec
x
d
x
.
Find the indefinite integral
∫
(
1
1
+
x
+
1
1
+
x
2
)
d
x
\int(\frac{1}{1+x}+\frac{1}{1+x^2})dx
∫
(
1
+
x
1
+
1
+
x
2
1
)
d
x
.
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Evaluate
∫
−
1
0
d
x
1
−
x
2
\int_{-1}^0\frac{dx}{\sqrt{1-x^2}}
∫
−
1
0
1
−
x
2
d
x
Antiderivatives: Inverse Trigonometric Functions
Find
∫
2
4
−
t
2
d
t
\int_{ }^{ }\frac{2}{\sqrt{4-t^2}}dt
∫
4
−
t
2
2
d
t
.
More Antiderivatives of Exponential Functions Questions:
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Practice: Definite Integral (~F2017 Final Q26)
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
Antiderivatives: Exponential Functions
Find the indefinite integral
∫
(
6
x
+
e
−
6
x
)
d
x
\displaystyle \int(6^x+e^{-6x}) \ dx
∫
(
6
x
+
e
−
6
x
)
d
x
Exponential Functions: $f$ from $f''$
Given
f
′
′
(
x
)
=
5
e
x
+
2
sin
x
f''(x)=5e^x+2\sin x
f
′′
(
x
)
=
5
e
x
+
2
sin
x
and given
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
,
f
(
π
)
=
0
f(\pi)=0
f
(
π
)
=
0
, find
f
(
x
)
f(x)
f
(
x
)
. Enter "f(x)=..."
Indefinite Integrals with Exponential
Compute the following integral:
∫
3
−
x
d
x
\displaystyle \int 3^{-x}\,\text{d}x
∫
3
−
x
d
x
.
Use upper case
C
C
C
to denote any constants.
Antiderivatives: Exponential Functions
Which of the following is the most general antiderivative of the function
e
3
x
+
7
e^{3x+7}
e
3
x
+
7
? In the functions below,
c
c
c
is an arbitrary constant.
Find
∫
(
c
o
s
7
)
e
3
x
d
x
\displaystyle \int (cos7)e^{3x}dx
∫
(
cos
7
)
e
3
x
d
x
Antiderivatives: Exponential Functions
Find an antiderivative of
f
(
x
)
=
2
e
3
x
−
2
f(x) = 2e^{3x-2}
f
(
x
)
=
2
e
3
x
−
2
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\displaystyle \int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
Find the equation of the tangent line of the graph
y
=
e
x
cos
x
y=\frac{e^x}{\cos\ x}
y
=
c
o
s
x
e
x
at the point where 𝑥 = 0.
Exponential Functions: $f$ from $f''$
Given
f
′
′
(
x
)
=
5
e
x
+
2
sin
x
f''(x)=5e^x+2\sin x
f
′′
(
x
)
=
5
e
x
+
2
sin
x
and given
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
,
f
(
π
)
=
0
f(\pi)=0
f
(
π
)
=
0
, find
f
(
x
)
f(x)
f
(
x
)
. Enter "f(x)=..."
Indefinite Integrals with Exponential
Compute the following integral:
∫
3
−
x
d
x
\displaystyle \int 3^{-x}\,\text{d}x
∫
3
−
x
d
x
.
Use upper case
C
C
C
to denote any constants.
Antiderivatives: Exponential Functions
Find the indefinite integral
∫
(
6
x
+
e
−
6
x
)
d
x
\displaystyle \int(6^x+e^{-6x}) \ dx
∫
(
6
x
+
e
−
6
x
)
d
x
Find the equation of the tangent line to the curve
y
=
2
3
e
3
−
3
x
y=2^3e^{3-3x}
y
=
2
3
e
3
−
3
x
at
x
=
1
x=1
x
=
1
.
Enter your answer as y=...
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Higher Order Derivatives
Find a general formula for the
n
th-derivative of
f
(
x
)
=
e
3
x
f(x)=e^{3x}
f
(
x
)
=
e
3
x
.
Antiderivatives: Trigonometric and Exponential Functions
If
f
′
′
(
x
)
=
cos
(
2
x
)
−
x
4
+
e
x
f''\left(x\right)=\cos\left(2x\right)-x^4+e^x
f
′′
(
x
)
=
cos
(
2
x
)
−
x
4
+
e
x
,
f
(
0
)
=
0
=
f
′
(
0
)
f\left(0\right)=0=f'\left(0\right)
f
(
0
)
=
0
=
f
′
(
0
)