MHB Fundamental theorem of calculus and more....

[ertagon2]

So as always I come here to make sure my maths homework is right and ask few questions to make sure I understand the topic.
Here is my homework:
View attachment 7705

Q.1 I'm fairly certain that this is correct, however, please check if I didn't do any stupid mistakes.

Q.2 Same as above.

Q.3 Now here is where the problems start is the equation inside the integral sign f(x)'? What about "c"? Does it cancel out? How? I know this is based on fundamental theorem of calculus can you explain using normal language what it is. Is the answer correct ?

Q.4 I differentiated the equation. Inserted 4 as x. Got 5.11... Is this correct? Same questions as Q.3

Q.5 Seems easy enough. Just basic calculus. Please check if right.

 

[skeeter]

#2 and #4 are incorrect ... the rest are ok

 

[ertagon2]

#2 and #4 are incorrect ... the rest are ok

I just checked and it seems that the answer to Q.2 is 4.5. Could you please explain what is incorrect?

 

[MountEvariste]

Q.2 Same as above.

It's incorrect. I think you have found $\displaystyle \int_{-2}^{1}|1+2x| \,{dx}$ which is different to $\displaystyle \int_{-2}^{1}1+2x \,{dx}$.

Q.3 Now here is where the problems start is the equation inside the integral sign f(x)'? What about "c"? Does it cancel out? How? I know this is based on fundamental theorem of calculus can you explain using normal language what it is. Is the answer correct ?

Yes, the answer is correct. But I'll try to answer some of your queries regarding this part.

The part of theorem is they want you to use is: if $\displaystyle F(x) = \int_a^x\!f(t)\,d{t}$ then $F'(x) = f(x)$. So, the equation inside integral is not $f(x)$. It's $f(t)$.

It's a definite integral, not an indefinite integral. The bound/limit variable $x$ defines your function, whereas in indefinite integrals your function is defined in terms of a variable in the integrand. In Plain English, if your function $F(x)$ is defined by an integral of $f(t)$ (with respect to $t$) going from a constant (in this case $a$) to a variable (in this case $x$), then you can find the derivative of your function $F(x)$ by taking $f(t)$ and replacing $t$ with $x.$

If you're wondering 'what happened to the $a$?' 'whereas the $C$?' etc and are thinking of in terms of indefinite integrals, perhaps the following example might prove useful. Consider the following, where have $f(t) = t^2$.

$\displaystyle F(x) = \int_2^{x} t^2\,{dt}$

Suppose want to find $F'(x)$. First, if you find the indefinite integral you get:

$\displaystyle \int t^2 \,{dt} = \frac{t^3}{3}+C$

But remember this is not $F(x)$. To find $F(x)$, put the limits in

$\displaystyle \frac{t^3}{3}+C \bigg|_{t=2}^{t=x} = \bigg(\frac{x^3}{3}+C\bigg)-\bigg(\frac{2^3}{3}+C\bigg)$

And the $C$'s cancel so we're left with $\displaystyle \frac{x^3}{3}-\frac{2^3}{3}~ $ $\Longleftarrow $ and this is $F(x)$.

Now that we have $F(x)$, to find $F'(x)$ we just differentiate: $\displaystyle F'(x) = \frac{d}{dx} \bigg(\frac{x^3}{3}-\frac{2^3}{3}\bigg) = x^2.$

So the part that resulted from the lower bound constant $a=2$ disappeared after we differentiated because it was a constant.

The statement if $\displaystyle F(x) = \int_a^x\!f(t)\,d{t}$ then $F'(x) = f(x)$ shortens this process (since the part given by the lower bound is constant and differentiates to $0$).

Instead of going all through that we look at the integral and say $f(t) = t^2$ so $f(x) = x^2$ hence $F'(x) = f(x) = x^2.$There's another part of the theorem you may have learned that says if $F'(x) = f(x)$

$\displaystyle \int_a^b f(t)\, dt = F(b)-F(a)$.

Q.4 I differentiated the equation. Inserted 4 as x. Got 5.11... Is this correct? Same questions as Q.3

The answer is not correct. You may have missed/misplaced a constant. The method is the same as above.

 

[ertagon2]

It's incorrect. I think you have found $\displaystyle \int_{-2}^{1}|1+2x| \,{dx}$ which is different to $\displaystyle \int_{-2}^{1}1+2x \,{dx}$. Yes, the answer is correct. But I'll try to answer some of your queries regarding this part.

The part of theorem is they want you to use is: if $\displaystyle F(x) = \int_a^x\!f(t)\,d{t}$ then $F'(x) = f(x)$. So, the equation inside integral is not $f(x)$. It's $f(t)$.

It's a definite integral, not an indefinite integral. The bound/limit variable $x$ defines your function, whereas in indefinite integrals your function is defined in terms of a variable in the integrand. In Plain English, if your function $F(x)$ is defined by an integral of $f(t)$ (with respect to $t$) going from a constant (in this case $a$) to a variable (in this case $x$), then you can find the derivative of your function $F(x)$ by taking $f(t)$ and replacing $t$ with $x.$

If you're wondering 'what happened to the $a$?' 'whereas the $C$?' etc and are thinking of in terms of indefinite integrals, perhaps the following example might prove useful. Consider the following, where have $f(t) = t^2$.

$\displaystyle F(x) = \int_2^{x} t^2\,{dt}$

Suppose want to find $F'(x)$. First, if you find the indefinite integral you get:

$\displaystyle \int t^2 \,{dt} = \frac{t^3}{3}+C$

But remember this is not $F(x)$. To find $F(x)$, put the limits in

$\displaystyle \frac{t^3}{3}+C \bigg|_{t=2}^{t=x} = \bigg(\frac{x^3}{3}+C\bigg)-\bigg(\frac{2^3}{3}+C\bigg)$

And the $C$'s cancel so we're left with $\displaystyle \frac{x^3}{3}-\frac{2^3}{3}~ $ $\Longleftarrow $ and this is $F(x)$.

Now that we have $F(x)$, to find $F'(x)$ we just differentiate: $\displaystyle F'(x) = \frac{d}{dx} \bigg(\frac{x^3}{3}-\frac{2^3}{3}\bigg) = x^2.$

So the part that resulted from the lower bound constant $a=2$ disappeared after we differentiated because it was a constant.

The statement if $\displaystyle F(x) = \int_a^x\!f(t)\,d{t}$ then $F'(x) = f(x)$ shortens this process (since the part given by the lower bound is constant and differentiates to $0$).

Instead of going all through that we look at the integral and say $f(t) = t^2$ so $f(x) = x^2$ hence $F'(x) = f(x) = x^2.$There's another part of the theorem you may have learned that says if $F'(x) = f(x)$

$\displaystyle \int_a^b f(t)\, dt = F(b)-F(a)$.
The answer is not correct. You may have missed/misplaced a constant. The method is the same as above.

Thank you very much. It made everything so much easier.
About Q.2 I didn't read it correctly. I thought they wanted me to calculate the area of triangles e.g $\displaystyle \int_{-2}^{1}|1+2x| \,{dx}$ but they only want me to use geometric approach to find the net area under/above function e.g $\displaystyle \int_{-2}^{1}1+2x \,{dx}$ which can be also done using normal calculus. The answer is "0" am I right ?

 

[MountEvariste]

Thank you very much. It made everything so much easier.
About Q.2 I didn't read it correctly. I thought they wanted me to calculate the area of triangles e.g $\displaystyle \int_{-2}^{1}|1+2x| \,{dx}$ but they only want me to use geometric approach to find the net area under/above function e.g $\displaystyle \int_{-2}^{1}1+2x \,{dx}$ which can be also done using normal calculus. The answer is "0" am I right ?

Spot on!

 

[MarkFL]

...About Q.2 I didn't read it correctly. I thought they wanted me to calculate the area of triangles e.g $\displaystyle \int_{-2}^{1}|1+2x| \,{dx}$ but they only want me to use geometric approach to find the net area under/above function e.g $\displaystyle \int_{-2}^{1}1+2x \,{dx}$ which can be also done using normal calculus. The answer is "0" am I right ?

You could also use a substitution here to check your result:

$$u=x+\frac{1}{2}\implies du=dx$$

And the definite integral becomes:

$$2\int_{-\Large{\frac{3}{2}}}^{\Large{\frac{3}{2}}} u\,du$$

And by the odd-function rule, we know this evaluates to zero.

 

[ertagon2]

It's incorrect. I think you have found $\displaystyle \int_{-2}^{1}|1+2x| \,{dx}$ which is different to $\displaystyle \int_{-2}^{1}1+2x \,{dx}$. Yes, the answer is correct. But I'll try to answer some of your queries regarding this part.

The part of theorem is they want you to use is: if $\displaystyle F(x) = \int_a^x\!f(t)\,d{t}$ then $F'(x) = f(x)$. So, the equation inside integral is not $f(x)$. It's $f(t)$.

It's a definite integral, not an indefinite integral. The bound/limit variable $x$ defines your function, whereas in indefinite integrals your function is defined in terms of a variable in the integrand. In Plain English, if your function $F(x)$ is defined by an integral of $f(t)$ (with respect to $t$) going from a constant (in this case $a$) to a variable (in this case $x$), then you can find the derivative of your function $F(x)$ by taking $f(t)$ and replacing $t$ with $x.$

If you're wondering 'what happened to the $a$?' 'whereas the $C$?' etc and are thinking of in terms of indefinite integrals, perhaps the following example might prove useful. Consider the following, where have $f(t) = t^2$.

$\displaystyle F(x) = \int_2^{x} t^2\,{dt}$

Suppose want to find $F'(x)$. First, if you find the indefinite integral you get:

$\displaystyle \int t^2 \,{dt} = \frac{t^3}{3}+C$

But remember this is not $F(x)$. To find $F(x)$, put the limits in

$\displaystyle \frac{t^3}{3}+C \bigg|_{t=2}^{t=x} = \bigg(\frac{x^3}{3}+C\bigg)-\bigg(\frac{2^3}{3}+C\bigg)$

And the $C$'s cancel so we're left with $\displaystyle \frac{x^3}{3}-\frac{2^3}{3}~ $ $\Longleftarrow $ and this is $F(x)$.

Now that we have $F(x)$, to find $F'(x)$ we just differentiate: $\displaystyle F'(x) = \frac{d}{dx} \bigg(\frac{x^3}{3}-\frac{2^3}{3}\bigg) = x^2.$

So the part that resulted from the lower bound constant $a=2$ disappeared after we differentiated because it was a constant.

The statement if $\displaystyle F(x) = \int_a^x\!f(t)\,d{t}$ then $F'(x) = f(x)$ shortens this process (since the part given by the lower bound is constant and differentiates to $0$).

Instead of going all through that we look at the integral and say $f(t) = t^2$ so $f(x) = x^2$ hence $F'(x) = f(x) = x^2.$There's another part of the theorem you may have learned that says if $F'(x) = f(x)$

$\displaystyle \int_a^b f(t)\, dt = F(b)-F(a)$.
The answer is not correct. You may have missed/misplaced a constant. The method is the same as above.

Is Q.4 in radians ? I got "8" as an answer when I used radians instead of degreed. If so how was I supposed to know that I should use radians

Q.4

$\frac{f(x)''= cos(\pi{x}^{2})\pi2x}{\pi}$

how was I supposed to know that cos() is in radians ? I mean it could be as well number of degrees based on $\pi$

 

[Greg]

In calculus, we assume radians. If degrees are used they should be specified but that is rare if not non-existent.

 

[HOI]

The derivation of $\dfrac{d \sin(x)}{dx}= \cos(x)$ uses the limit $\displaystyle\lim_{x\to 0}\dfrac{\sin(x)}{x}= 1$. That is true only if $x$ is in radians.