Two values for $\int_1^4 \frac{2+x^2}{\sqrt{x}} dx$?

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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 $\frac{152}{3} - \frac{7}{3} = \frac{145}{3}$.