Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $G(x) = \begin{cases} x(t-1), & \text{where } x \leq t \\ t(x-1), & \text{where } t < x \end{cases}$ and if $f$ is continuous function of $x$ in $[0, 1]$. Let $g(x) = \int_0^1 f(t)G(x,t)dt$, then:
g(0) = 1
g(0) = 0
g(1) = 1
g''(x) = f(x)
Step-by-Step Solution
Key Concept: Recognizing that the derivative of $\tan^{-1}x$ appears in the integrand allows us to use substitution to reduce the problem to an integration by parts.
We rewrite the integrand using $\cot^{-1}x$ properties: $\int e^{\tan^{-1}x}(1+x+x^2)/(\cot^{-1}x) dx$ becomes $\int e^{\tan^{-1}x}(1+x^2)\frac{1}{1+x^2}dx$ after substitution $\tan^{-1}x = t$, so $dx = \frac{1}{1+x^2}dt$. This simplifies to $-\int e^t(\tan t + \sec^2 t)dt = -e^t \tan t + c = -xe^{\tan^{-1}x} + c$.
Correct Answer: 2,3