Improper Integrals - AP Calculus BC
Card 0 of 132
Evaluate:

Evaluate:
First, we will find the indefinite integral
using integration by parts.
We will let
and
.
Then
and
.






To find
, we use another integration by parts:
, which means that
, and
, which means that, again,
.











Since
, or,

for all real
, and
,
by the Squeeze Theorem,
.





First, we will find the indefinite integral using integration by parts.
We will let and
.
Then and
.
To find , we use another integration by parts:
, which means that
, and
, which means that, again,
.
Since
, or,
for all real , and
,
by the Squeeze Theorem,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral,
.
We will let
and
.
Then,
and
.

and







Now, this expression evaluated at
is equal to
.
At
it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as
, as follows:

and
, so by L'Hospital's rule,

Therefore,


![=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right ]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/180705/gif.latex)


First, we will find the indefinite integral, .
We will let and
.
Then,
and
.
and
Now, this expression evaluated at is equal to
.
At it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as , as follows:
and
, so by L'Hospital's rule,
Therefore,
Compare your answer with the correct one above
Evaluate:

Evaluate:
Rewrite the integral as
.
Substitute
. Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so


Since
, the above is equal to
.
Rewrite the integral as
.
Substitute . Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so
Since , the above is equal to
.
Compare your answer with the correct one above
Evaluate
.
Evaluate .
By the Formula Rule, we know that
. We therefore know that
.
Continuing the calculation:



By the Power Rule for Integrals,
for all
with an arbitrary constant of integration
. Therefore:



.
So, 
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382522/gif.latex)
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382523/gif.latex)
.


By the Formula Rule, we know that . We therefore know that
.
Continuing the calculation:
By the Power Rule for Integrals, for all
with an arbitrary constant of integration
. Therefore:
.
So,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral
using integration by parts.
We will let
and
.
Then
and
.






To find
, we use another integration by parts:
, which means that
, and
, which means that, again,
.











Since
, or,

for all real
, and
,
by the Squeeze Theorem,
.





First, we will find the indefinite integral using integration by parts.
We will let and
.
Then and
.
To find , we use another integration by parts:
, which means that
, and
, which means that, again,
.
Since
, or,
for all real , and
,
by the Squeeze Theorem,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral,
.
We will let
and
.
Then,
and
.

and







Now, this expression evaluated at
is equal to
.
At
it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as
, as follows:

and
, so by L'Hospital's rule,

Therefore,


![=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right ]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/180705/gif.latex)


First, we will find the indefinite integral, .
We will let and
.
Then,
and
.
and
Now, this expression evaluated at is equal to
.
At it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as , as follows:
and
, so by L'Hospital's rule,
Therefore,
Compare your answer with the correct one above
Evaluate:

Evaluate:
Rewrite the integral as
.
Substitute
. Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so


Since
, the above is equal to
.
Rewrite the integral as
.
Substitute . Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so
Since , the above is equal to
.
Compare your answer with the correct one above
Evaluate
.
Evaluate .
By the Formula Rule, we know that
. We therefore know that
.
Continuing the calculation:



By the Power Rule for Integrals,
for all
with an arbitrary constant of integration
. Therefore:



.
So, 
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382522/gif.latex)
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382523/gif.latex)
.


By the Formula Rule, we know that . We therefore know that
.
Continuing the calculation:
By the Power Rule for Integrals, for all
with an arbitrary constant of integration
. Therefore:
.
So,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral
using integration by parts.
We will let
and
.
Then
and
.






To find
, we use another integration by parts:
, which means that
, and
, which means that, again,
.











Since
, or,

for all real
, and
,
by the Squeeze Theorem,
.





First, we will find the indefinite integral using integration by parts.
We will let and
.
Then and
.
To find , we use another integration by parts:
, which means that
, and
, which means that, again,
.
Since
, or,
for all real , and
,
by the Squeeze Theorem,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral,
.
We will let
and
.
Then,
and
.

and







Now, this expression evaluated at
is equal to
.
At
it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as
, as follows:

and
, so by L'Hospital's rule,

Therefore,


![=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right ]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/180705/gif.latex)


First, we will find the indefinite integral, .
We will let and
.
Then,
and
.
and
Now, this expression evaluated at is equal to
.
At it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as , as follows:
and
, so by L'Hospital's rule,
Therefore,
Compare your answer with the correct one above
Evaluate:

Evaluate:
Rewrite the integral as
.
Substitute
. Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so


Since
, the above is equal to
.
Rewrite the integral as
.
Substitute . Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so
Since , the above is equal to
.
Compare your answer with the correct one above
Evaluate
.
Evaluate .
By the Formula Rule, we know that
. We therefore know that
.
Continuing the calculation:



By the Power Rule for Integrals,
for all
with an arbitrary constant of integration
. Therefore:



.
So, 
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382522/gif.latex)
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382523/gif.latex)
.


By the Formula Rule, we know that . We therefore know that
.
Continuing the calculation:
By the Power Rule for Integrals, for all
with an arbitrary constant of integration
. Therefore:
.
So,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral
using integration by parts.
We will let
and
.
Then
and
.






To find
, we use another integration by parts:
, which means that
, and
, which means that, again,
.











Since
, or,

for all real
, and
,
by the Squeeze Theorem,
.





First, we will find the indefinite integral using integration by parts.
We will let and
.
Then and
.
To find , we use another integration by parts:
, which means that
, and
, which means that, again,
.
Since
, or,
for all real , and
,
by the Squeeze Theorem,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral,
.
We will let
and
.
Then,
and
.

and







Now, this expression evaluated at
is equal to
.
At
it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as
, as follows:

and
, so by L'Hospital's rule,

Therefore,


![=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right ]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/180705/gif.latex)


First, we will find the indefinite integral, .
We will let and
.
Then,
and
.
and
Now, this expression evaluated at is equal to
.
At it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as , as follows:
and
, so by L'Hospital's rule,
Therefore,
Compare your answer with the correct one above
Evaluate:

Evaluate:
Rewrite the integral as
.
Substitute
. Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so


Since
, the above is equal to
.
Rewrite the integral as
.
Substitute . Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so
Since , the above is equal to
.
Compare your answer with the correct one above
Evaluate
.
Evaluate .
By the Formula Rule, we know that
. We therefore know that
.
Continuing the calculation:



By the Power Rule for Integrals,
for all
with an arbitrary constant of integration
. Therefore:



.
So, 
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382522/gif.latex)
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382523/gif.latex)
.


By the Formula Rule, we know that . We therefore know that
.
Continuing the calculation:
By the Power Rule for Integrals, for all
with an arbitrary constant of integration
. Therefore:
.
So,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral
using integration by parts.
We will let
and
.
Then
and
.






To find
, we use another integration by parts:
, which means that
, and
, which means that, again,
.











Since
, or,

for all real
, and
,
by the Squeeze Theorem,
.





First, we will find the indefinite integral using integration by parts.
We will let and
.
Then and
.
To find , we use another integration by parts:
, which means that
, and
, which means that, again,
.
Since
, or,
for all real , and
,
by the Squeeze Theorem,
.
Compare your answer with the correct one above
Evaluate:

Evaluate:
First, we will find the indefinite integral,
.
We will let
and
.
Then,
and
.

and







Now, this expression evaluated at
is equal to
.
At
it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as
, as follows:

and
, so by L'Hospital's rule,

Therefore,


![=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right ]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/180705/gif.latex)


First, we will find the indefinite integral, .
We will let and
.
Then,
and
.
and
Now, this expression evaluated at is equal to
.
At it is undefined, because
does not exist.
We can use L'Hospital's rule to find its limit as , as follows:
and
, so by L'Hospital's rule,
Therefore,
Compare your answer with the correct one above
Evaluate:

Evaluate:
Rewrite the integral as
.
Substitute
. Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so


Since
, the above is equal to
.
Rewrite the integral as
.
Substitute . Then
and
. The lower bound of integration stays
, and the upper bound becomes
, so
Since , the above is equal to
.
Compare your answer with the correct one above
Evaluate
.
Evaluate .
By the Formula Rule, we know that
. We therefore know that
.
Continuing the calculation:



By the Power Rule for Integrals,
for all
with an arbitrary constant of integration
. Therefore:



.
So, 
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382522/gif.latex)
![=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/382523/gif.latex)
.


By the Formula Rule, we know that . We therefore know that
.
Continuing the calculation:
By the Power Rule for Integrals, for all
with an arbitrary constant of integration
. Therefore:
.
So,
.
Compare your answer with the correct one above