1.

Find a + 3b if\(\smallint \left( {{x^2} - 1} \right)\;dx = \frac{{{x^3}}}{a} + bx + C\)where C is integration constant ?1. 22. 03. 14. None of these

Answer» Correct Answer - Option 2 : 0

CONCEPT:

  • \(\smallint {x^n}\;dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C\)
  • \(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

CALCULATION:

Given:\(\smallint \left( {{x^2} - 1} \right)\;dx = \frac{{{x^3}}}{a} + bx + C\)

As we know that,\(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

\(\smallint \left( {{x^2} - 1} \right)\;dx = \smallint {x^2}\;dx - \smallint dx \)

As we know that,\(\smallint {x^n}\;dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C\)

\(\smallint \left( {{x^2} - 1} \right)\;dx = \frac{{{x^3}}}{3} - x + C\)

Now, by comparing the above equation with\(\smallint \left( {{x^2} - 1} \right)\;dx = \frac{{{x^3}}}{a} + bx + C\)we get

⇒ a = 3 and b = - 1

⇒ a + 3b = 0

Hence, correct option is 2.



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