I have worked a definite integral . . .

$$\int\limits_1^4 \frac{2+x^2}{\sqrt{x}} dx\\[2em] \int\limits_1^4 \frac{2+x^2}{1} \cdot \frac{1}{\sqrt{x}} \;dx\\[2em] \left. \frac{2x}{1} +\frac{x^3}{3}\cdot\frac{x^{1/2}}{1} \;\right|^4_1\\[2em] \left. \frac{2x}{1} +\frac{x^{7/2}}{3} \;\right|^4_1\\[2em] \left. \frac{x^{7/2} + 6x}{3}\;\right|^4_1\\[2em] \left. \frac{(x^3)^{1/2} + 6x}{3}\;\right|^4_1\\[2em] 32+24\\[2em] \frac{56}{3}-\frac{7}{3} = \frac{49}{3} $$

As you can see, I got the solution $\frac{49}{3}$ or $16.3333333...$

The correct answer in the textbook is $\frac{82}{5}$ or $16.4$

Is it possible for a definite integral to have slightly different answers? If so, why? If not, where exactly are my calculations incorrect?

I've gone over this multiple times and still can't figure it out. Thanks for any help provided.

  • $\begingroup$ You can't do that! You forgot to integrate the product! $\endgroup$
    – Andrew Li
    Apr 24, 2018 at 2:15
  • 5
    $\begingroup$ It was a good effort, but you broke up the integrand incorrectly. If you would like to practice more problems just like this one, take a look at lem.ma/CV $\endgroup$
    – Lemma
    Apr 24, 2018 at 2:38
  • $\begingroup$ You could notice something was wrong when the original problem has a degree of $3/2$ and your integrated expression ends up with a degree of $7/2$. $\endgroup$
    – dan post
    Apr 24, 2018 at 4:05

3 Answers 3


No it is not. You made an error - you wrote this: $$\int \frac{2+x^2}1\cdot\frac{1}{\sqrt x}\,dx= \int \left( \frac{2+x^2}1+\frac{1}{\sqrt x}\right)\,dx$$, then you integrated this, then you recombined the product. This is where your mistake is. It is merely a coincident that the two answers are so close.

If you do it correctly:

$$\int \frac{2+x^2}1\cdot\frac{1}{\sqrt x}\,dx=\int\left(\frac2{\sqrt x}+x^{3/2}\right)\,dx$$ ...then you should get the correct answer given in your textbook.

  • 2
    $\begingroup$ That's not the only error; they also integrate $x^2 \cdot \frac{1}{\sqrt x}$ as $\int x^2 \int \frac 1 {\sqrt x}$. $\endgroup$
    – user296602
    Apr 24, 2018 at 2:17
  • $\begingroup$ @T.Bongers Ah yes thanks, I didn't notice that. Added $\endgroup$
    – John Doe
    Apr 24, 2018 at 2:19
  • $\begingroup$ " then you integrated this" Technically, that can't be integrated; you can't integrate something without a infinitesimal term (dx), and the the 2+x^2 part is written as not being multiplied by the dx. $\endgroup$ Apr 24, 2018 at 3:27
  • $\begingroup$ @Acccumulation if you want rigor, you should say that you can't integrate something that isn't a differential form. $\endgroup$
    – rafa11111
    Apr 24, 2018 at 12:59

There are several errors in this calculation.

  • Line 2: What happened to $\mathrm{d}x$?
  • Line 3: The antiderivative of a product is not the product of antiderivatives. This is evident on considering $\int x \cdot x \,\mathrm{d}x$, which is $\frac{1}{3}x^3 + C$, NOT $\require{cancel}\bcancel{\frac{1}{4}x^4 + C}$.
  • Line 6: $x^{7/2} = x^3 x^{1/2}$, but is not $(x^3)^{1/2} = x^{3/2}$. (Caught by T. Bongers.)
  • Line 7: $4^{7/2} = 128$, so assuming that there had been no prior errors, you should have $\frac{152}{3} - \frac{7}{3} = \frac{145}{3}$.
  • $\begingroup$ Maybe one to add / modify: Line 6 involves writing $x^{7/2}$ as $(x^3)^{1/2}$. $\endgroup$
    – user296602
    Apr 24, 2018 at 2:25
  • 1
    $\begingroup$ @T.Bongers : Yowza. Didn't notice that in the image. Added. $\endgroup$ Apr 24, 2018 at 2:29
  • $\begingroup$ I mostly added the comment because I thought that it explained the line 7 issue... but then I realized that $32 = 4^{5/2}$. I guess the exponent goes down by one on each line?? $\endgroup$
    – user296602
    Apr 24, 2018 at 2:34

Your integration is wrong. Hint: You cannot integrate $\frac{2+x^2}{1}$ and $\frac{1}{\sqrt{x}}$ separately.


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