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

Answer
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363.6k+ views
Hint: Evaluate the given integral carefully without missing any term in between.

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

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.
Last updated date: 21st Sep 2023
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