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I am asked to find the area between ${y = 7}$ and ${x^2 -5x + 13}$

Combining these equations together I get ${-x^2 - 5x + 6 = 0}$.

Factorising into ${(x - 3)(x - 2)}$

I am taking ${y = 7}$ to be the top equation so I then find the integral by subtracting the first equation from the second:

$${\int_{2}^3 (7 - (x^2 -5x + 13))\ dx}$$

Am I right so far?

This becomes:

$${\int_{2}^3 (-x^2 +5x - 6})dx$$

The integral is:

${-{x^3\over 3} + {5x^2\over 2} -6x}$

I subtract the 2 domain values:

$${({-27\over 3} + {45\over 2} - 18 ) - (-{8\over 3} + 10 - 12)}$$

$\implies {5{1\over 2} - {-4 {2\over 3}}}$

$\implies {10 {1\over 6}}$

This answer is wrong by some distance and I just cannot see what I am doing wrong.

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  • $\begingroup$ Check the line after you say, "This becomes: ..." $\endgroup$
    – fosho
    Feb 5, 2016 at 18:55
  • $\begingroup$ From your first integral to your second integral, doesn't $7 - (x^2-5x+13)$ become $-x^2+5x-6$? $\endgroup$
    – Maximilian Gerhardt
    Feb 5, 2016 at 18:56
  • $\begingroup$ Two issues: first, what you wrote immediately after subtracting has a sign error, but you corrected it further down. Second, your arithmetic for the final answer is wrong: wolframalpha.com/input/… $\endgroup$
    – Ian
    Feb 5, 2016 at 18:56
  • $\begingroup$ The area on what interval? $\endgroup$
    – mvw
    Feb 5, 2016 at 19:03
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    $\begingroup$ Note that $-\dfrac{27}{3}+\dfrac{45}{2}-18=-\dfrac{9}{2}.$ $\endgroup$
    – mfl
    Feb 5, 2016 at 19:08

2 Answers 2

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Notice, your method is correct but you have made a mistake while subtracting in the last $$\left(\frac{-27}{3}+\frac{45}{2}-18\right)-\left(\frac{-8}{3}+10-12\right)$$ $$=\left(\frac{-9}{2}\right)-\left(\frac{-14}{3}\right)$$ $$=\frac{-9}{2}+\frac{14}{3}$$ $$=\frac{-27+28}{6}=\color{red}{\frac 16}$$

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Here is an alternate method:

Plot the line: $y=7$ & upward parabola: $y=x^2-5x+13$ or $\left(x-\frac 52\right)^2=y-\frac 14$ which are intersecting each other at the points $\left(2, 7\right)$ & $(3, 7)$. Indicate the area between the line & the curve which is enclosed in a rectangle of $7\times 1$ & is given as $$(\text{area of rectangle})-(\text{area under the curve:}\ x^2-5x+13 \ \text{between}\ x=2\ \ \text{&} \ \ x=3)$$
$$=7\times 1-\int_2^3 (x^2-5x+13)\ dx$$ $$=7-\left[\frac{x^3}{3}-\frac {5x^2}{2}+13x\right]_2^3 $$ $$=7-\frac{41}{6}=\color{red}{\frac 16}$$

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  • $\begingroup$ why is the first equation outside the integral, I.e 7 - ${\int}$ $\endgroup$
    – dagda1
    Feb 5, 2016 at 19:37
  • $\begingroup$ This an alternate method by considering a rectangle enclosing the required area. The required area is equal to the difference of area of rectangle of $7\times 1$ to the area under the curve $x^2-5x+13$ between $x=2$ & $x=3$ which is given as $7-\int$ $\endgroup$
    – Harish Chandra Rajpoot
    Feb 5, 2016 at 19:46

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