Question Number 122075 by peter frank last updated on 13/Nov/20
Answered by MJS_new last updated on 14/Nov/20
![seriously??? =∫_0 ^1 ∣x+1∣dx=∫_0 ^1 (x+1)dx=[(x^2 /2)+x]_0 ^1 =(3/2)](https://www.tinkutara.com/question/Q122078.png)
$$\mathrm{seriously}??? \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mid{x}+\mathrm{1}\mid{dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({x}+\mathrm{1}\right){dx}=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by peter frank last updated on 14/Nov/20
$$\mathrm{thank}\:\mathrm{you} \\ $$