CA Foundation · Quantitative Aptitude · Differential and Integral Calculus
If ∫ from 0 to k of (2x + 3) dx = 10 and k > 0, then the value of k is:
Integrating 2x + 3 gives x² + 3x, so the definite integral from 0 to k equals k² + 3k. Setting this to 10 gives k² + 3k − 10 = 0, whose positive root is k = 2.
- A1
- B2Correct
- C3
- D5
Explanation
The integral equals [x² + 3x] from 0 to k = k² + 3k. Setting k² + 3k = 10 gives k² + 3k − 10 = 0, so (k + 5)(k − 2) = 0. Since k > 0, k = 2. Check: 4 + 6 = 10. The root −5 is rejected because k is positive.
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