Question

evaluate 2x-1 dx from 0 to 4​

Answer 1

$\displaystyle \sf \int\limits_{0}^{4} (2x - 1) \, dx = 12$

Given :

$\displaystyle \sf \int\limits_{0}^{4} (2x - 1) \, dx$

To find :

Integrate the integral

Solution :

Step 1 of 2 :

Write down the given Integral

The given Integral is

$\displaystyle \sf \int\limits_{0}^{4} (2x - 1) \, dx$

Step 2 of 2 :

Integrate the integral

$\displaystyle \sf \int\limits_{0}^{4} (2x - 1) \, dx$

$\displaystyle \sf = \bigg[ 2. \frac{ {x}^{2} }{2} - x \bigg]_{0}^{4}$

$\displaystyle \sf = \bigg[ {x}^{2} - x \bigg]_{0}^{4}$

$\displaystyle \sf = \bigg[ {4}^{2} - 4 \bigg] - \bigg[ {0}^{2} - 0 \bigg]$

$\displaystyle \sf = \bigg[ 16 - 4 \bigg] - \bigg[ 0 - 0 \bigg]$

$\displaystyle \sf = 12$

Correct question : Integrate 2x-1 dx from 0 to 4