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I was thinking if i could calculate the integral of a sum ... and i started by a simple function ... f(x)=x; and to make it a little harder i tried to solve for a; And here is the integral. ... integral (from 0 to a) of the sum (from x=1 to a) of x. All is equal to 6. Here are the steps i passed through to conclude that "a" is equal to 2. my calculation of the integral So i want you to let me know if i did it correctly or no ?... and if it is correct ... then what is it's meaning ? (The area of a SUM ?? ) Thank you for your help.
PS : "a" is an integer.

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    To take the integral, you need to know what the sum with a non-integer upper bound means. – 2017-02-01
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    At least on my screen the text in the image is rotated and too light to be legible. Please see the FAQ about MathJax – 2017-02-01
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    Sorry i forgot to mention that a is an integer – 2017-02-01
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    I edited my post thanks – 2017-02-01
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    A cubic equation has three roots, not one. – 2017-02-01
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    First we are working only in the integers – 2017-02-01
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    And secondly i just plugged the numbers from 0 to 5 and found that 2 is the answer – 2017-02-01

1 Answers 1

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Your work is almost correct. That is, upto the integration and summation.

However, you must remember that a cubic equation has 3 roots, not just one. So the last step needs to be modified and the equation will be as follows:

$$a^3+a^2-12=0$$ $$(a-2)(a^2+3a+6)=0$$

So the roots of the quadratic equation are: $$a=\frac{-3\pm\sqrt{9-4\times1\times 6}}{2\cdot 1}$$ $$a=\frac{-3\pm i\sqrt{15}}{2}$$

So we get 2 complex conjugate roots. But since it is given that $a$ is an integer, so these two complex roots are not possible. $a = 2$ is the only answer.

Hope this helps you.