MHB Calculating Integrals with Antiderivatives

[mathmari]

Hey!

Let $a, b \in \mathbb{R}$ with $a<b$ and $f\in C^1([a,b])$.

I want to calculate the following integrals using the fundamental theorem of calculus, by finding in each case an antiderivative.

1. $\int_a^b\frac{f'(x)}{f(x)}$ with $0\notin f([a,b])$
   
2. $\int_a^bf'(x)f(x)dx$
   
3. $\int_a^b\frac{f'(x)}{\sqrt{f(x)^2-1}}dx$ with $f([a,b])\subseteq (1, \infty)$
   
4. $\int_0^{\frac{\pi}{4}}\left (\frac{2}{\cos^2(x)}+4x\right )e^{\tan (x)+x^2}dx$

For the first two I have done the following:

1. We have that $\left (\ln (f(x))\right )'=\frac{f'(x)}{f(x)}$, so an antiderivative is $F(x)=\ln (f(x))$.
   
   So, we have that $$\int_a^b\frac{f'(x)}{f(x)}=F(b)-F(a)=\ln (f(b))-\ln (f(a))=\ln \left (\frac{f(b)}{f(a)}\right )$$
   
2. We have that $\left (f^2(x)\right )'=2f(x)f'(x) \Rightarrow f(x)f'(x)=\frac{1}{2}\left (f^2(x)\right )'$, so an antiderivatice is $F(x)=\frac{1}{2}f^2(x)$.
   
   So, we have that $$\int_a^bf'(x)f(x)dx=F(b)-F(a)=\frac{1}{2}\left (f^2(b)-f^2(a)\right )$$

Is everything correct? (Wondering) For 3. if we had the integral $\int_a^b\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}dx$, an antiderivative would be $\sqrt{f^2(x)-1}$.

What could we do know where don't have the term $f(x)$ in numerator? (Wondering) Could you give me a hint for 4. ? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

For 3. if we had the integral $\int_a^b\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}dx$, an antiderivative would be $\sqrt{f^2(x)-1}$.

What could we do know where don't have the term $f(x)$ in numerator?

It's all correct.
I think that the best we can do without $f(x)$ in the numerator is to try partial integration.
It still means that $f(x)$ has to be somewhere outside of the square root though. (Thinking)

Could you give me a hint for 4. ?

What's the derivative for the part with the exponential function? (Wondering)

 

[mathmari]

I think that the best we can do without $f(x)$ in the numerator is to try partial integration.
It still means that $f(x)$ has to be somewhere outside of the square root though. (Thinking)

We have the following:
$$\int_a^b\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b-\int_a^bf(x)\left ( \frac{1}{\sqrt{f^2(x)-1}}\right )'dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b-\int_a^bf(x)\left ( (f^2(x)-1)^{-\frac{1}{2}}\right )'dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b-\int_a^bf(x)\left ( -\frac{1}{2}(f^2(x)-1)^{-\frac{3}{2}}2f(x)f'(x)\right )dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b+\int_a^bf(x)(f^2(x)-1)^{-\frac{3}{2}}f(x)f'(x)dx$$

Does this help us? (Wondering)

 

[Prove It]

Hey!

Let $a, b \in \mathbb{R}$ with $a<b$ and $f\in C^1([a,b])$.

I want to calculate the following integrals using the fundamental theorem of calculus, by finding in each case an antiderivative.

1. $\int_a^b\frac{f'(x)}{f(x)}$ with $0\notin f([a,b])$
   
2. $\int_a^bf'(x)f(x)dx$
   
3. $\int_a^b\frac{f'(x)}{\sqrt{f(x)^2-1}}dx$ with $f([a,b])\subseteq (1, \infty)$
   
4. $\int_0^{\frac{\pi}{4}}\left (\frac{2}{\cos^2(x)}+4x\right )e^{\tan (x)+x^2}dx$

For the first two I have done the following:

1. We have that $\left (\ln (f(x))\right )'=\frac{f'(x)}{f(x)}$, so an antiderivative is $F(x)=\ln (f(x))$.
   
   So, we have that $$\int_a^b\frac{f'(x)}{f(x)}=F(b)-F(a)=\ln (f(b))-\ln (f(a))=\ln \left (\frac{f(b)}{f(a)}\right )$$
   
2. We have that $\left (f^2(x)\right )'=2f(x)f'(x) \Rightarrow f(x)f'(x)=\frac{1}{2}\left (f^2(x)\right )'$, so an antiderivatice is $F(x)=\frac{1}{2}f^2(x)$.
   
   So, we have that $$\int_a^bf'(x)f(x)dx=F(b)-F(a)=\frac{1}{2}\left (f^2(b)-f^2(a)\right )$$

Is everything correct? (Wondering) For 3. if we had the integral $\int_a^b\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}dx$, an antiderivative would be $\sqrt{f^2(x)-1}$.

What could we do know where don't have the term $f(x)$ in numerator? (Wondering) Could you give me a hint for 4. ? (Wondering)

3.

$\displaystyle \begin{align*} \int_a^b{ \frac{f'(x)}{\sqrt{\left[ f(x) \right] ^2 - 1}}\,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} f(x) = \cosh{(t)} \implies f'(x)\,\mathrm{d}x = \sinh{(t)}\,\mathrm{d}t \end{align*}$, nothing that when $\displaystyle \begin{align*} x = a, \, t = \textrm{arcosh}\,\left[ f(a) \right] \end{align*}$ and when $\displaystyle \begin{align*} x = b, \, t = \textrm{arcosh}\,\left[ f(b) \right] \end{align*}$ to find

$\displaystyle \begin{align*} \int_a^b{ \frac{f'(x)}{\sqrt{ \left[ f(x) \right] ^2 - 1 }}\,\mathrm{d}x } &= \int_{\textrm{arcosh}\,\left[ f(a) \right]}^{\textrm{arcosh}\,\left[ f(b) \right] }{ \frac{\sinh{(t)}}{\sqrt{\cosh^2{(t)} - 1}} \,\mathrm{d}t } \\ &= \int_{\textrm{arcosh}\,\left[ f(a) \right]}^{\textrm{arcosh}\,\left[ f(b) \right]}{ \frac{\sinh{(t)}}{\sinh{(t)}}\,\mathrm{d}t } \\ &= \int_{\textrm{arcosh}\,\left[ f(a) \right] }^{\textrm{arcosh}\,\left[ f(b)\right] }{ 1\,\mathrm{d}t } \\ &= \left[ t \right] _{\textrm{arcosh}\,\left[ f(a) \right] }^{\textrm{arcosh}\,\left[ f(b) \right] } \\ &= \textrm{arcosh}\,\left[ f(b) \right] - \textrm{arcosh}\,\left[ f(a) \right] \end{align*}$4.

$\displaystyle \begin{align*} \int_0^{\frac{\pi}{4}}{ \left[ \frac{2}{\cos^2{(x)}} + 4\,x \right] \,\mathrm{e}^{ \tan{(x)} + x^2 }\,\mathrm{d}x } &= 2\int_0^{\frac{\pi}{4}}{ \left[ \frac{1}{\cos^2{(x)}} + 2\,x \right] \,\mathrm{e}^{\tan{(x)} + x^2} \,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} u = \tan{(x)} + x^2 \implies \mathrm{d}u = \left[ \frac{1}{\cos^2{(x)}} + 2\,x \right] \,\mathrm{d}x \end{align*}$ and note that when $\displaystyle \begin{align*} x = 0 , \, u = 0 \end{align*}$ and when $\displaystyle \begin{align*} x = \frac{\pi}{4} , \, u = 1 + \frac{\pi ^2}{16} \end{align*}$ to find

$\displaystyle \begin{align*} 2 \int_0^{ \frac{\pi}{4} }{ \left[ \frac{1}{\cos^2{(x)}} + 2\,x \right] \,\mathrm{e}^{\tan{(x)} + x^2} \,\mathrm{d}x } &= 2 \int_0^{ 1 + \frac{\pi ^2}{16} }{ \mathrm{e}^{u}\,\mathrm{d}u } \\ &= 2 \, \left[ \mathrm{e}^u \right]_0^{ 1 + \frac{\pi ^2}{16} } \\ &= 2\,\left[ \mathrm{e}^{ 1 + \frac{\pi^2}{16} } - \mathrm{e}^0 \right] \\ &= 2 \,\left[ \mathrm{e}^{ 1 + \frac{\pi ^2}{16} } - 1 \right] \end{align*}$

 

[mathmari]

Ahh I see! (Nerd)

Now where we use the substitution, can we say which the antiderivative in each case is? (Wondering)

 

[Prove It]

Ahh I see! (Nerd)

Now where we use the substitution, can we say which the antiderivative in each case is? (Wondering)

Yes of course you can, but it is ALWAYS more concise to change the bounds when you do the substitution to avoid having to do this.

In 3. for example we ended up with t + C, and as we made the substitution $\displaystyle \begin{align*} f(x) = \cosh{(t)} \end{align*}$ that means $\displaystyle \begin{align*} t = \textrm{arcosh}\,\left[ f(x) \right] \end{align*}$ and thus the antiderivative is $\displaystyle \begin{align*} \textrm{arcosh}\,\left[ f(x) \right] + C \end{align*}$.

 

[I like Serena]

We have the following:
$$\int_a^b\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b-\int_a^bf(x)\left ( \frac{1}{\sqrt{f^2(x)-1}}\right )'dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b-\int_a^bf(x)\left ( (f^2(x)-1)^{-\frac{1}{2}}\right )'dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b-\int_a^bf(x)\left ( -\frac{1}{2}(f^2(x)-1)^{-\frac{3}{2}}2f(x)f'(x)\right )dx=\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}\mid_a^b+\int_a^bf(x)(f^2(x)-1)^{-\frac{3}{2}}f(x)f'(x)dx$$

Does this help us? (Wondering)

Not to solve this particular integral, but if we had to solve the integral that you've just found, we could use partial integration to bring it back to the original integral.

To actually solve the integral, we can do a substitution, which is effectively what you did:
$$\int_a^b\frac{f'(x)f(x)}{\sqrt{f(x)^2-1}}dx
= \int_a^b\frac{f(x)}{\sqrt{f(x)^2-1}}df(x)
=\int_{f(a)}^{f(b)}\frac{u}{\sqrt{u^2-1}}du
$$