- #1

chwala

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- 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...

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...