Mathematics, 07.08.2021 04:20 Tamiahufflululu
Yet another Calculus question...
This time, it's about integrals/antiderivatives. The question asks to find the value of ![\int\limits^6_0 {[g(x)+2]} \, dx](/tpl/images/2435/3681/2fac8.png)
. Since the derivative of 2 is 0, can I directly find 
? Thanks in advance!
Answers: 3
![\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 32](/tpl/images/1405/6104/14c6b.png)
![\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx](/tpl/images/1405/6104/4bd0c.png)
![\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx](/tpl/images/1405/6104/e6b91.png)
![\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = \int\limits^6_0 {g(x)} \, dx + \int\limits^6_0 {2} \, dx](/tpl/images/1405/6104/6be2a.png)
[2nd Integral] Rewrite [Integration Property - Multiplied Constant]: ![\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = \int\limits^6_0 {g(x)} \, dx + 2\int\limits^6_0 {} \, dx](/tpl/images/1405/6104/2b9d1.png)
[1st Integral] Substitute in value: ![\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 20 + 2\int\limits^6_0 {} \, dx](/tpl/images/1405/6104/8971f.png)
[Integral] Reverse Power Rule: ![\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 20 + 2(x) \bigg| \limits^6_0](/tpl/images/1405/6104/9954c.png)
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: ![\displaystyle \int\limits^6_0 {[g(x) + 2]} \, dx = 20 + 2(6)](/tpl/images/1405/6104/94be3.png)
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Yet another Calculus question...
This time, it's about integrals/antiderivatives. The question asks...
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