1. ## integral question

I am studying for my final and I have come across two questions from an old exam that I can not figure out. Thanks to anyone who is willing to lend an ear!

integral of f(x) dx + 2 equals integral of f(x) dx + 6 both from 2 to 5

I am forgetting how to figure this out.

integral of sin^2 t dt I have tried by substitution, by parts with no luck

the answer on test is 1/2 integral of (1 - cos(2t)) dt How do you know this??????

Thanks again, I LOVE and APPRECIATE this site more than I can say!

2. Is this it: $\displaystyle \int_{2}^{5} f(x) dx + 2 = \int_{2}^{5} f(x) dx + 6$

It's impossible. Let $\displaystyle a = \int_{2}^{5} f(x)dx$:
$\displaystyle a + 2 = a + 6 \: \: \Rightarrow \: \: 2 = 6$ an absurd conclusion.

--------------------------------------------

$\displaystyle \int \sin^{2}t \: dt$

Use the identity: $\displaystyle \sin^{2} \theta = \frac{1 - \cos (2\theta)}{2}$

3. ## integral question

the first integral should be

from 2 to 5 of (f(x) + 2) dx

while the second one is from 2 to 5 of f(x) dx + 6

Sorry that I didn't make that clear before.

4. $\displaystyle \int_2^5 [f(x)+2]~dx$

$\displaystyle \int_2^5 f(x)~dx~+~\int_2^5 2~dx$

$\displaystyle \int_2^5 f(x)~dx~+~2x \bigg |_2^5$

$\displaystyle \int_2^5 f(x)~dx + 6$

5. Originally Posted by Frostking
the first integral should be

from 2 to 5 of (f(x) + 2) dx

while the second one is from 2 to 5 of f(x) dx + 6

Sorry that I didn't make that clear before.
So we have

$\displaystyle \int_2^5 \left(f(x) + 2\right)dx = \int_2^5f(x)\,dx + 6$,

correct? And I suppose you are trying to show that this statement is true? You can do:

$\displaystyle \int_2^5 \left(f(x) + 2\right)dx = \int_2^5f(x)\,dx + 6$

$\displaystyle \Rightarrow\int_2^5f(x)\,dx + \int_2^52\,dx = \int_2^5f(x)\,dx + 6$

$\displaystyle \Rightarrow\int_2^52\,dx = 6$

$\displaystyle \Rightarrow2x\bigg|_2^5 = 6\Rightarrow2(5 - 2) = 6\Rightarrow6 = 6$

And the left side is equal to the right, provided that $\displaystyle f(x)$ is indeed integrable over that interval.