Question

evaluate integral of x/2+3 dx with limit -2 to 4 ​

Answer 1

Explanation:

The ans for this question is 21

Answer 2

$\implies \displaystyle \sf\int\limits^{4}_{ - 2} \dfrac{x}{2} + 3 \, \: dx$

We know that :-

$\underline{\displaystyle \bf \boxed{\displaystyle \sf\int\limits^{a}_{ b} f(x) \, \: dx =g(x)]^{a} _{b} = g(a) - g(b)}}$

$\implies \displaystyle \sf\int\limits^{4}_{ - 2} \dfrac{x}{2} + 3 \, \: dx = \dfrac{ {x}^{2} }{4} + 3 x]^{ + 4} _{ - 2}$

Now put the value of upper limit and lower limit.

$\implies \sf\dfrac{ {4}^{2} }{4} + (3 \times 4)] - \dfrac{ { (- 2)}^{2} }{4} + (3 \times - 2)$

$\implies \sf {4}+ 12 - (1 - 6)$

$\implies \sf 16- ( - 5)$

$\implies \sf 16 + 5$

$\implies \sf21$

Answer :

$\displaystyle \sf\int\limits^{4}_{ - 2} \dfrac{x}{2} + 3 \, \: dx = 21$