<s>[INST] Integration and Limits I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.
$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$
Please help out. Thanks.
 [/INST] The integration is obtained as follows:
$$\int 2\frac{x+2}{5}dx=\frac{2}{5}\int (x+2)d(x+2)=\frac{2}{5}\int udu=\frac{2}{5}\frac{u^2}{2}=\frac 1 5 (x+2)^2$$
Since $\frac 1 5 (x+2)^2$ is a primitive of $2\frac{x+2}{5}$ we can use FTCII, and get
$$\int 2\frac{x+2}{5}dx=\frac{(\color{red}{1}+2)^2}{5}-\frac{(\color{red}{0}+2)^2}{5}= \frac 9 5- \frac 4 5 = 1$$
It seems what you did was this:
$$\int 2\frac{x+2}{5}dx=\frac{(0+2)^2}{5}-\frac{(0+1)^2}{5}= \frac 4 5- \frac 1 5 = \frac 3 5$$
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