# Integration and Limits

I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.

$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$

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:

$$\int 2\frac{x+2}{5}dx=\frac{(0+2)^2}{5}-\frac{(0+1)^2}{5}= \frac 4 5- \frac 1 5 = \frac 3 5$$

• Would not we be integrating x separately and 2 separately? – Fahad Uddin Jun 19 '12 at 22:59
• @StartupCrazy See the edit. – Pedro Tamaroff Jun 19 '12 at 23:04
• Thanks a lot. I had actually forgotten to multiply the 2 outside with the second term. – Fahad Uddin Jun 19 '12 at 23:09

$$\left. \dfrac{(x+2)^2}5 \right \vert_0^1 = \dfrac{(1+2)^2}5 - \dfrac{(0+2)^2}5 = \dfrac{3^2}5 - \dfrac{2^2}5 = \dfrac{9}5 - \dfrac45 = \dfrac{9-4}5 = \dfrac55 = 1$$

Note that we can integrate $\displaystyle \int_{x_1}^{x_2} (x+a) dx$ in seemingly two different ways.

The first method is to treat $x+a$ together as one object i.e. $$\displaystyle \int (x+a) dx = \dfrac{(x+a)^2}2 + c_1$$

The second method is to treat $x+a$ as two separate objects i.e. $$\displaystyle \int (x+a) dx = \int x dx + \int a dx = \dfrac{x^2}2 + ax + c_2$$

It might seem that both are different. However, note that the first answer can be re-written as $$\dfrac{(x+a)^2}2 + c_1 = \dfrac{x^2}2 + ax + \dfrac{a^2}2 + c_1.$$ Now this looks more closely like the second. The only difference in fact is that the constants are different. They are in fact related as $c_2 = c_1 + \dfrac{a^2}2$. While performing a definite integral, the constants cancel off and hence both ways should give us the same answer.

As an exercise, we will integrate what you have by treating $x$ and $2$ separately.

\begin{align} \int_0^1 \dfrac{2(x+2)}5 dx & = \dfrac25 \int_0^1 (x+2)dx = \dfrac25 \int_0^1 xdx + \dfrac25 \int_0^1 2dx = \dfrac25 \cdot \left. \dfrac{x^2}2 \right \vert_{0}^1 + \dfrac25 \cdot 2 \cdot \left(1 - 0 \right)\\ & = \dfrac25 \cdot \left(\dfrac{1^2}2 - \dfrac{0^2}2 \right) + \dfrac45 = \dfrac25 \cdot \dfrac12 + \dfrac45 = \dfrac15 + \dfrac45 = 1 \end{align}

• The downvote did get you going! Undone. – Pedro Tamaroff Jun 19 '12 at 23:15
• @PeterTamaroff Actually nope. I wrote the detailed answer and figured out only in the end that there was a down vote :). Also, in future, it would be great to leave a comment if you decide when you down vote. – user17762 Jun 19 '12 at 23:17
• I usually do, but I thought I'd be evident. – Pedro Tamaroff Jun 19 '12 at 23:19

$$\left.\frac{(x+2)^2}{5}\right|_0^1 = \frac{(1+2)^2}{5} - \frac{(0+2)^2}{5} = \frac{9}{5} - \frac{4}{5} = \frac{5}{5} = 1$$

• Would not we be integrating x separately and 2 separately? – Fahad Uddin Jun 19 '12 at 22:59
• @StartupCrazy - I don't understand your question. What do you mean by integrating $2$? – Ayman Hourieh Jun 19 '12 at 23:03
• Why would you integrate $x$ and $2$ separately? You need to understand how to use the fundamental theorem of calculus for integration : If you find a function $F(x)$ such that $F'(x) = f(x)$, then $\int_a^b f(x) \, dx = F(b) - F(a)$. In this case, the function $\frac{(x+2)^2}5$ is such that its derivative is precisely $\frac{2(x+2)}5$, hence you can evaluate the integral by computing $\frac{(x+2)^2}5$ evaluated at $1$ minus the same thing evaluated at $0$. – Patrick Da Silva Jun 19 '12 at 23:03
• If you want to use the fact that the integral is linear, then you can compute $$\int_0^1 \frac{2(x+2)}5 \,dx = \int_0^1 \frac {2x}5 \, dx + \int_0^1 \frac {4}5 \, dx = \left. (\frac{x^2}5 + \frac {4x}5) \right|_0^1,$$ and the only difference between $\frac{x^2}5 + \frac 45$ and $\frac{(x+2)^2}5$ is perhaps a constant, but it will cancel itself out when you compute $F(b) - F(a)$. – Patrick Da Silva Jun 19 '12 at 23:04
• Oh, Thankyou. Yes, I was applying linearity on it but did a mistake. Thanks a lot. – Fahad Uddin Jun 19 '12 at 23:07

Here's another potential approach that you will likely find useful in the future (though it isn't really necessary, here), called "$u$-substitution".

Let's put $u=x+2$. Now, $x=0$ if and only if $u=2$, and $x=1$ if and only if $u=3$. Also, $$\frac{du}{dx}=\frac{d}{dx}[x+2]=1,$$ and if we treat $\frac{du}{dx}$ just like any other fraction and "solve" for $dx$, we get $du=dx$. Now we'll go through and substitute everything $x$-related with the corresponding $u$-related thing, so that $$\int_0^1\frac{2(x+2)}{5}\,dx=\int_2^3\frac{2u}{5}\,du=\left.\frac{u^2}{5}\right|_2^3=1.$$

Now, $u$-substitutions typically require a bit more finagling than this one did, but as a preview (and an alternate approach), I think it serves its purpose.