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Question: If the value of \[\int_1^k {(2x - 3)dx = 12} \], then find the value of k. \[({\text{a}})\] \[ - 2...

If the value of ∫1k(2x−3)dx=12\int_1^k {(2x - 3)dx = 12} , then find the value of k.
(a)({\text{a}}) −2 - 2 and5{\text{5}}
(b){\text{(b)}} 5{\text{5}}and 2{\text{2}}
(c){\text{(c)}} 2{\text{2}} and - 5{\text{ - 5}}
(d){\text{(d)}}None of these

Explanation

Solution

Hint: Evaluate the given integral carefully without missing any term in between.

We have the given integral as
∫1k(2k−3)dx\int_1^k {(2k - 3)dx}
After integrating the above equation, we get,
=[x2−3x]1k= [{x^2} - 3x]_1^k
=(k2−3k)−(1−3)= ({k^2} - 3k) - (1 - 3)
=k2−3k+2= {k^2} - 3k + 2
According to the question,
We are given that the value of the given integral is equal to 1212,
Therefore, we get
k2−3k+2=12{k^2} - 3k + 2 = 12
k2−3k−10=0{k^2} - 3k - 10 = 0
This equation can be re written in the form as
k2−5k+2k−10=0{k^2} - 5k + 2k - 10 = 0
k(k−5)+2(k−5)=0k(k - 5) + 2(k - 5) = 0
(k+2)(k−5)=0(k + 2)(k - 5) = 0
∴k=−2,5\therefore k = - 2,5
Therefore, the required solution is (a)({\text{a}}) −2 - 2 and5{\text{5}}.

Note: In these types of questions, the given integral is solved, then equated to the values given in the question, which on evaluation gives the value of the required variable.