Logarithms!

Calculus Level 3

$\large \int_0^1 \ln x \, dx + \int_{-1}^0 \ln(x+1) \, dx = \, ?$

The answer is -2.

1 solution

A Former Brilliant Member
Jun 4, 2016

For $\int_{0}^{1} \ln x\,dx + \int _{-1}^{0} \ln \left(x+1\right)\,dx$

$\begin{aligned} \int_{0}^{1} \ln x\,dx + \int _{-1}^{0} \ln \left(x+1\right)\,dx &= [x \ln(x)-x \color{grey}{\ce{+C}}]_{0}^{1}+[-x+(x+1) \ln(x+1)-1 \color{grey}{+\ce{C}}]_{-1}^{0} \quad\quad\quad{\text{Indefinite*}}\\& =[-1-0]+[-1-0] \\&= -2 \space \square \end{aligned}$

*For clarification by computing definite integrals you have to remove the constant ( $+\ce{C}$ ) after you integrate to avoid confuse.

$\large \text{ADIOS!!!}$

Very nice. But how did you get your last second step?

Akash Shukla - 5 years ago

Use the definite formula

$\int_{a}^{b} f(x)\,dx = F(b) - F(a)$

$\displaystyle{\text{OK!}}$

A Former Brilliant Member - 5 years ago

Haha.. thank you. But by doing this we get 0*log(0), which is not defined

Akash Shukla - 5 years ago

@Akash Shukla – For the second step of $\ [x \ln(x)-x \color{grey}{\ce{+C}}]_{0}^{1}+[-x+(x+1) \ln(x+1)-1 \color{grey}{+\ce{C}}]_{-1}^{0}$ you have to remove the constant

were

$[x \ln(x)-x \color{grey}{\ce{+C}}]_{0}^{1}+[-x+(x+1) \ln(x+1)-1 \color{grey}{+\ce{C}}]_{-1}^{0}$

for

$\begin{aligned} [x \ln(x)-x]_{0}^{1} &=x \ln(x) \quad{\text{Apply L'Hôpital's Rule}} \\&= \left(\frac{\ln(x)}{\frac{1}{x}}\right)_{0}^{1} \\&= \left(\frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}\left(\frac{1}{x}\right)}\right)_{0}^{1} \\&= \left(\frac{\frac{1}{x}}{-\frac{1}{x^2}}\right)_{0}^{1} \quad\quad\quad\quad{\text{Reciprocal}} \\&= (-x)_{0}^{1} \quad\quad\quad\quad{\text{Substitute}} \\&=(-1) - 0 = -1 \end{aligned}$

and

$\begin{aligned}(-x+(x+1) \ln(x+1)-1)_{-1}^{0} &= ((0)+((0)+1) \ln((0)+1))-((-1)+((-1)+1)\ln((-1)+1)) \\&=-1\end{aligned}$

where $\left[x \ln(x)-x \color{grey}{\ce{+C}}\right]_{0}^{1}+\left[-x+(x+1) \ln(x+1)-1 \color{grey}{+\ce{C}}\right]_{-1}^{0} = -2$ $\square$

$\LARGE\text{ADIOS!!!}$

A Former Brilliant Member - 5 years ago

@A Former Brilliant Member – Thank you. Actually I don't know about this, so I have questioned on your solution.

Akash Shukla - 5 years ago

@Akash Shukla – Asking is a principal interaction of conversation.

A Former Brilliant Member - 5 years ago

@Akash Shukla – L'Hôpital's Rule rule helps solve that problem.

tytan le nguyen - 5 years ago

@Tytan Le Nguyen – I'm sorry I forgot using L'Hôpital's rule thank you for recalling that.

A Former Brilliant Member - 5 years ago

@Tytan Le Nguyen – That rule is useful on limits. But how to use that L'Hôpital's Rule on integration?

A Former Brilliant Member - 5 years ago

@Akash Shukla – Actually $0\ce{LOG}_{0}(x)=\infty$ .

A Former Brilliant Member - 5 years ago

@A Former Brilliant Member – I have done it by expansion of log. The method you suggested involves $0*log(0)$ , which is not defined.

Akash Shukla - 5 years ago

Aren't the two integrals the same value since they're the same functions, only one is translated left one unit (along with the limits of integration)? If that is true then you should just have to multiply the first one by 2 instead of writing out the entire second antiderivative.

Ryan Kelly - 5 years ago

Yes, they both are same

Akash Shukla - 5 years ago

Logarithms!

The answer is -2.

1 solution

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