Author Topic: Exponential and Logs Question (Read 1211 times) Tweet Share

georgebanis · « **on:** March 21, 2019, 07:20:47 pm »

Hey guys,

I have a couple of maths questions from the Cambridge Textbook attached if you could give me a hand.

Thanks

fun_jirachi · « **Reply #1 on:** March 21, 2019, 09:10:30 pm »

The main idea to take away from this is that integration is anti-differentiation, and if you like, 'every integral gives us another derivative' and vice versa (as said by my teacher.)

$\frac{d}{dx}x^2 \ln x=2x \ln x+x \ \text{(By the product rule)} \\ \text{Then,} \ \int_\sqrt{e}^{e} x \ln x \ dx = \frac{1}{2}\int_\sqrt{e}^{e} 2x \ln x + x - x \ dx \\=\frac{1}{2}\left[x^2 \ln x - \frac{x^2}{2}\right]_\sqrt{e}^{e} \\ \text{You should be able to take it from here.} \\ \frac{d}{dx}\sqrt{x} \ln x = \frac{\ln x}{2\sqrt{x}}+\sqrt{x} \ \text{(By the product rule)} \\ \text{Apply similar methods to the above question to find the indefinite integral,} \\ \text{and don't forget +C}$

Try the rest yourself, and if you don't understand, feel free to ask again! Hope this helps

Kombmail · « **Reply #2 on:** June 06, 2019, 09:41:14 am »

Jirachi! How does -2(t+1)^-1 / -1 when integrated become 2log e (t+1)?

DrDusk · « **Reply #3 on:** June 06, 2019, 05:03:53 pm »

Quote from: Kombmail on June 06, 2019, 09:41:14 am

Jirachi! How does -2(t+1)^-1 / -1 when integrated become 2log e (t+1)?

$\text{We have:} \\ f(t) = \frac{-2(t+1)^{-1}}{-1} \\ = \frac{2}{t+1}$
$= 2\int \frac{1}{t+1}\,dt \\ \text{The integral is in the form of} \int \frac{f'(t)}{f(t)}\,dt$
$\text{which is a standard integral that evaluates to} \\ ln(f(t)) + C \\ \text{thus giving:} \\ F(t) = 2ln(t+1) + C$

Kombmail · « **Reply #4 on:** June 06, 2019, 07:15:07 pm »

Quote from: DrDusk on June 06, 2019, 05:03:53 pm

$\text{We have:} \\ f(t) = \frac{-2(t+1)^{-1}}{-1} \\ = \frac{2}{t+1}$
$= 2\int \frac{1}{t+1}\,dt \\ \text{The integral is in the form of} \int \frac{f'(t)}{f(t)}\,dt$
$\text{which is a standard integral that evaluates to} \\ ln(f(t)) + C \\ \text{thus giving:} \\ F(t) = 2ln(t+1) + C$

Thanks!

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Author Topic: Exponential and Logs Question (Read 1211 times) Tweet Share

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