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The diagram shows part of a sketch of the curve with equation $y=\frac{2}{x^2}+x$. The points $A$ and $B$ have $x$-coordinates $\frac 12$ and $2$ respectively. Find the area of the finite region between $AB$ and the curve.

First, I found the $y$ values for $A$ and $B$ by substituting $x$ into the equation of the curve. I got $A(\frac 12, \frac{17}{2})$ and $B(2, \frac 52)$.

Then, I used the formula for the area of a trapezium: $\frac{a+b}{2} \times h$, to find the area under the line. $ \frac {\frac {17}{2} + \frac 52} {2} \times \frac 32 = \frac {33}{4}$ square units.

I then used $\int_{\frac 12}^2(y=\frac{2}{x^2}+x)dx$ to find the area under the curve. $[\frac {2x^{-1}}{-1} + x^2]_\frac 12^2 = \frac {27}{4}$ square units.

If I subtract the latter area from the former, I get the answer $\frac 32$ square units. The textbook states the answer is $\frac {27}{8}$. Why am I wrong?

MrRookie
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    How can we possibly know why you are wrong, if you don't show us your working? You don't tell us your $y$ values, you don't tell us the area of the trapezium, you don't tell us what you got for the area under the curve or how you got it. Any or all of these could be wrong. – TonyK Apr 21 '20 at 14:30
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    Maybe you should check your calculations? – Devansh Kamra Apr 21 '20 at 14:36
  • Apologies. I thought I was making a more fundamental mistake, as I checked over my working several times before posting the question. Question updated. – MrRookie Apr 21 '20 at 17:48
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    The antiderivative of $x$ is $\frac{1}{2}x^2$, not $x^2$. – Paul Apr 21 '20 at 17:48
  • Ah, don't know how I missed that. Cheers – MrRookie Apr 21 '20 at 17:51

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