Question

help with integrals

Answer 1

Evalute \[\int\limits_{(0,1)}^{(2,5)}\left( 3x+y \right)dx+\left( 2y-x \right)dy\] along the curve \[y=x ^{2}+1\]

Answer 2

Can't just integrate the first quantity with respect to x and the 2nd with respect to y, then plug in the limits of integration?

Answer 3

then what is the use for y =x^2 +1

Answer 4

The limits of integration are points along that curve.

Answer 5

:)
Let's parametrize the curve.
\[\Large x = t\]\[\Large y = t^2 +1\]

Answer 6

Now, your limits of integration are from t = 1 to t = 2

Answer 7

And then, let's differentiate both equations with respect to t.
\[\Large dx = dt\]\[\Large dy = 2tdt\]

Answer 8

Can you see where this is going?

Answer 9

yes ,i forgot how to do parametric can u hint me .

Answer 10

Well, it's basically a matter of substitution now, first, let's deal with
\[\Large x = t\]and \[\Large dx = dt\]
\[\Large\int\limits_{C}^{}\left( 3\color{red}t+y \right)\color{red}{dt}+\left( 2y-\color{red}t \right)dy\]

Catch me so far?

Answer 11

yep

Answer 12

Okay, now on to
\[\Large y = t^2 + 1\]
and \[\Large dy = 2t dt\]
\[\Large\int\limits_{C}^{}\left( 3\color{}t+\color{blue}{t^2+1}\right)\color{}{dt}+\left( 2\color{blue}{(t^2+1)}-\color{}t \right)\color{blue}{2tdt}\]

We can now factor out the dt, which gives us
\[\Large \int\limits_C^{}\left[(3t^2 + t^2 + 1)+2t\left(2(t^2+1)-t\right)\right]dt\]
Basically a polynomial :)

Answer 13

Whoops, sorry, I meant
\[\Large \int\limits_C^{}\left[(3t + t^2 + 1)+2t\left(2(t^2+1)-t\right)\right]dt\]

Answer 14

Can you take it from here? ^_^

Answer 15

yes thanks

Answer 16

Remember that your curve runs from t = 1 to t = 2
And it's integration 101 XD