Is the method used to evaluate the given integral correct?

[chwala]

Homework Statement

This is my own question (set by myself)..I am refreshing.

Evaluate the integral

$$\int_{-1}^0 |4t+2| dt$$

Relevant Equations

Fundamental theorem of calculus -definite integrals

Method 1,
Pretty straightforward,

$$\int_{-1}^0 |4t+2| dt$$

Let $u=4t+2$

$du=4 dt$

on substitution,

$$\frac{1}{4}\int_{-2}^2 |u| du=\frac{1}{4}\int_{-2}^0 (-u) du+\frac{1}{4}\int_{0}^2 u du=\frac{1}{4}[2+2]=1$$

Now on method 2,

$$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} |4t+2| dt+\int_{-0.5}^0 |4t+2| dt=(-0.5-0)+-0.5=|-1|=1$$

We take the absolute value when finding area under curves...

your insight welcome....this things need refreshing at all times... looks like the methods are just one and the same...

 

[scottdave]

I'm not sure what you mean by "we take the absolute value when finding area..."

The function is an absolute value. The methods are similar. How would this problem differ if there were no absolute value? A plot may help.

The u-substitution makes it intuitive where to split into 2 integrals.

 

[chwala]

I'm not sure what you mean by "we take the absolute value when finding area..."

The function is an absolute value. The methods are similar. How would this problem differ if there were no absolute value? A plot may help.

The u-substitution makes it intuitive where to split into 2 integrals.

@scottdave i was reffering to:

$$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} |4t+2| dt+\int_{-0.5}^0 |4t+2| dt=(-0.5-0)+-0.5=-1$$

...should the negative remain or it does not matter. This is a good question almost boggled me up trying to find the definite integral of $\int_{-1}^{-0.5} (4t+2) dt$, that is without the absolute value. $0!$ hmmmmm can't be ...

I later realized that the graph is split into two halves with $x=0.5$ as the point dividing them... to give us, Area under curve:
$A=0.5+0.5=1$ square units.

or does this follow the principle of odd and even functions where for instance for odd functions,
'$\int_{-a}^{a} f(x) dx=0$? ...really rusted in these area- i need to go through my notes!...

I think i got it, there is a difference between finding the definite integral and finding the area bound by the curve $y=f(x)$ having been given the limits say, $x_1$ and $x_2$.

 

[Mark44]

..should the negative remain or it does not matter.

You should not get a negative number. The graph of y = |4x + 2| is nonnegative for all real x, so any integral will also be nonnegative.

I later realized that the graph is split into two halves with x=0.5

You probably mean at x = -1/2.

 

[chwala]

You should not get a negative number. The graph of y = |4x + 2| is nonnegative for all real x, so any integral will also be nonnegative.

You probably mean at x = -1/2.

Yes at $x=-0.5$. I will check my working steps again...

Did you check my method $2$? that is in post $1$. Let me post the steps first. A minute.

$$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} |4t+2| dt+\int_{-0.5}^0 |4t+2|
dt=\left[2t^2+2t\right]_{-1}^{-0.5}+\left[2t^2+2t\right]_{-0.5}^{0}$$
....one needs to be quite clear on whether you're determining the definite integral or finding the area bound by the curve.

In our case, we are just evaluating the definite integral therefore,$$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} |4t+2| dt+\int_{-0.5}^0 |4t+2|
dt=\left[2t^2+2t\right]_{-1}^{-0.5}+\left[2t^2+2t\right]_{-0.5}^{0}$$

$$ =(-0.5-0)+(0+0.5)=-0.5+0.5=0$$

if it was area bound by the curve then our solution would be,$$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} |4t+2| dt+\int_{-0.5}^0 |4t+2|
dt=\left[2t^2+2t\right]_{-1}^{-0.5}+\left[2t^2+2t\right]_{-0.5}^{0}$$

$$ =(-0.5-0)+(0+0.5)=|-0.5|+0.5=1$$

 

[Mark44]

$$\int_{-1}^0 |4t+2| dt$$
Relevant Equations: Fundamental theorem of calculus -definite integrals

You can leave this section blank if there are no relevant equations. Fundamental Thm of Calculus is really too generic to be a helpful relevant equation.

Now on method 2, $$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} |4t+2| dt+\int_{-0.5}^0 |4t+2| dt=(-0.5-0)+-0.5=|-1|=1$$

I wouldn't use substitution (your method 1) for such a straightforward problem. I'm not saying it's wrong, just that I wouldn't go this route.

For your method 2, use the definition of absolute value to replace |4t + 2| by -4t - 2, when $t \le -1/2$. You have a mistake in your work, so you had to "fudge" your answer by taking the absolute value.

The integral you showed in your method 2 should look like this:
$$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} -4t - 2 ~dt+\int_{-0.5}^0 4t + 2 ~dt = \left . -2t^2 - 2t \right |_{-1}^{-1/2} + \left . 2t^2 + 2t \right |_{-1/2}^0$$
$$= -2/4 + 1 - 0 + 0 - (1/2 - 1) = 1$$
No fudging of the result was needed.

 

[chwala]

You can leave this section blank if there are no relevant equations. Fundamental Thm of Calculus is really too generic to be a helpful relevant equation.

I wouldn't use substitution (your method 1) for such a straightforward problem. I'm not saying it's wrong, just that I wouldn't go this route.

For your method 2, use the definition of absolute value to replace |4t + 2| by -4t - 2, when $t \le -1/2$. You have a mistake in your work, so you had to "fudge" your answer by taking the absolute value.

The integral you showed in your method 2 should look like this:
$$\int_{-1}^0 |4t+2| dt=\int_{-1}^{-0.5} -4t - 2 ~dt+\int_{-0.5}^0 4t + 2 ~dt = \left . -2t^2 - 2t \right |_{-1}^{-1/2} + \left . 2t^2 + 2t \right |_{-1/2}^0 $$
$$= -2/4 + 1 - 0 + 0 - (1/2 - 1) = 1$$
No fudging of the result was needed.

True, i should have taken the negative of the absolute value by considering $x=-0.5$...Noted.

On the side i think the question simply wanted us to evaluate the integral...check...

 

[Mark44]

Noted but i think the question simply wanted us to evaluate the integral...check...

Right, I get that. What I'm saying is that your work in method 2 was incorrect. You got the right answer in this method only by "fudging" your result.