# Thread: integration and another qn.

1. ## integration and another qn.

Find $\displaystyle \int\sin^2 (3x+2) dx$

and also another qn which i am not sure it belong to which catagory so i just post it here together with the previous qn.

the cubic polynomial is such that the coefficient of $\displaystyle x^3$is -2 ans the roots of the eqn f(x)=0 are 0.5,2 and k

Given that f(x) has a remainder -20 when divided by (x+2), find the value of k.

thank you

2. Originally Posted by helloying
the cubic polynomial is such that the coefficient of [tex]x^3[\math]is -2 ans the roots of the eqn f(x)=0 are 0.5,2 and k
Let the cubic polynomial,$\displaystyle f(x)=-2(x-0.5)(x-2)(x-k)$

Given that f(x) has a remainder -20 when divided by (x+2),
$\displaystyle f(-2)=-20$
$\displaystyle -2(-2-0.5)(-2-2)(-2-k)=-20$
which gives k=-3

3. Originally Posted by helloying

the cubic polynomial is such that the coefficient of [tex]x^3[\math]is -2 ans the roots of the eqn f(x)=0 are 0.5,2 and k

Given that f(x) has a remainder -20 when divided by (x+2), find the value of k.

thank you

.

4. Originally Posted by malaygoel
Let the cubic polynomial,$\displaystyle f(x)=0.2(x-0.5)(x-2)(x-k)$

$\displaystyle f(-2)=-20$
$\displaystyle 0.2(-2-0.5)(-2-2)(-2-k)=-20$
which gives k=8.

shouldn't it be f(x)=-2(x-0.5)(x-2)(x-k) as the leading coefficient is -2 .

shouldn't it be f(x)=-2(x-0.5)(x-2)(x-k) as the leading coefficient is -2 .
I corrected it...thanks...see now...is it correct????

6. thanks for your help. however i dont understand one thing. the qn mentioned that coefficent is -2 for x^3 but it did not mention for others.what if there are other coeficent for x^2 for something?will it have a differnet ans?

And also kindly help me solve the 1st qn i asked. Thanks

7. Originally Posted by helloying
Find $\displaystyle \int\sin^2 (3x+2) dx$

substitute$\displaystyle 3x+2=t$

8. Originally Posted by malaygoel
substitute$\displaystyle 3x+2=t$

em i dont know how to intergete sin^2. can u teach me how to do?

9. Originally Posted by helloying
em i dont know how to intergete sin^2. can u teach me how to do?
$\displaystyle \sin^{2}u = \frac{1}{2} - \frac{\cos 2u}{2}$