JEE Main 2024 — Definite Integration Question with Solution

To determine the integral of the piecewise function $$f$$ over the interval $$[-2,2]$$, we first ensure that $$f$$ is differentiable on $$\mathbb{R}$$, as given in the problem statement. Differentiability implies continuity, so $$f$$ must also be continuous at $$x=1$$.

The condition for continuity at $$x=1$$ is:

$$x^2+3x+a=bx+2$$ at $$x=1$$.

This simplifies to:

$$1+3+a=b(1)+2$$

$$\Rightarrow a=b-2$$

The first derivative of $$f$$ gives us two different expressions depending on the value of $$x$$:

For $$x\le 1$$, $$f^{\prime }(x)=2x+3$$; and for $$x>1$$, $$f^{\prime }(x)=b$$.

For $$f$$ to be differentiable at $$x=1$$, these derivatives must be equal at that point. Setting $$f^{\prime }(1)$$ from both expressions equal to each other gives $$2(1)+3=b$$, thus $$b=5$$. And from $$a=b-2$$, we have $$a=3$$.

Now, we can calculate the integral of $$f$$ over the specified interval:

$$\underset{-2}{\overset{1}{\int }}(x^2+3x+3)dx+\underset{1}{\overset{2}{\int }}(5x+2)dx$$

$$\begin{array}{rl}& ={\left[\frac{x^3}{3}+\frac{3x^2}{2}+3x\right]}_{-2}^1+{\left[\frac{5x^2}{2}+2x\right]}_1^2\\ & \\ & =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right)\\ & \\ & =6+\frac{3}{2}+12-\frac{5}{2}=17\end{array}$$