Question

You can help me?
$\sf \: find \: the\:answer \: of : \\ \bf \to \int^{1}_{ - 1} \: (2x + 1)dx = . \: . \: .$
​

Answer 1

~Hello! I will try to help you!

$\: \Rightarrow\sf \int^{1}_{ - 1} \: (2x + 1)dx = . \: . \: .$

»Our first step is to find the integration (2x + 1), then:-

$\Rightarrow \sf ( \frac{2}{1 + 1} {x}^{1 + 1} ) + x \: ]^{1}_{ - 1} \\ \Rightarrow \sf( \cancel{\frac{2}{2} } {x}^{2} ) + x \: ]^{1}_{ - 1} = \rm {x}^{2} + x \: ]^{1}_{ - 1}$

»The next step is to just substitute for the upper limit of 1 and the lower limit of -1 to find the result.

$\Rightarrow \sf( {1}^{2} + 1) - (( - 1) {}^{2} + ( - 1)) \\ \Rightarrow \sf(1 + 1) - (1 + ( - 1)){}{} \\ \Rightarrow \sf2 - 0 \\ \Rightarrow \bf2$

With this it's done!

Answer 2

Definite Integration

$\rightarrow \sf \int\limits^1_{-1} {(2x+1)} \, dx$

integrate the following

$\rightarrow \sf { [ \dfrac{2x^2}{2} +x ] }_{-1}^1$

simplify

$\rightarrow \sf [x^2+x]_{-1}^1$

apply limits

$\rightarrow \sf (1)^2+(1) - ( (-1)^2 +(-1))$

simplify

$\rightarrow \sf 2 - 0$

$\rightarrow \sf 2$