Question

Let \[ f(t)=\int_{0}^{t} e^{x^2}\Big((1+2x^2)\sin x+x\cos x\Big)\,dx. \] Then the value of \(f(\pi)-f\!\left(\frac{\pi}{2}\right)\) is equal to:

• A. \(-\pi e^{\pi^2/4}\)
• B. \(-\dfrac{\pi}{2}e^{\pi^2/4}\)
• C. \(\dfrac{\pi}{2}e^{\pi^2/4}\)
• D. \(\pi e^{\pi^2/4}\)

Hint:

Whenever you see \(e^{x^2}\) multiplied by algebraic–trigonometric terms, try expressing the integrand as the derivative of \(e^{x^2}\) times a simple function.

Answer 1

The Correct Option is C

Step 1: Observe the integrand Consider \[ e^{x^2}\Big((1+2x^2)\sin x+x\cos x\Big). \] We check if this is the derivative of a product: \[ \frac{d}{dx}\big(e^{x^2}\sin x\big) = e^{x^2}(2x\sin x+\cos x). \] Multiply this derivative by \(x\) and adjust: \[ \frac{d}{dx}\big(xe^{x^2}\sin x\big) = e^{x^2}\sin x + xe^{x^2}(2x\sin x+\cos x) \] \[ = e^{x^2}\big((1+2x^2)\sin x+x\cos x\big). \] Hence, \[ e^{x^2}\Big((1+2x^2)\sin x+x\cos x\Big) =\frac{d}{dx}\big(xe^{x^2}\sin x\big). \] Step 2: Evaluate \(f(t)\) \[ f(t)=\int_{0}^{t}\frac{d}{dx}\big(xe^{x^2}\sin x\big)\,dx \] \[ f(t)=\left[xe^{x^2}\sin x\right]_{0}^{t} =te^{t^2}\sin t. \] Step 3: Compute the required value \[ f(\pi)=\pi e^{\pi^2}\sin\pi=0 \] \[ f\!\left(\frac{\pi}{2}\right) =\frac{\pi}{2}e^{\pi^2/4}\sin\frac{\pi}{2} =\frac{\pi}{2}e^{\pi^2/4}. \] \[ f(\pi)-f\!\left(\frac{\pi}{2}\right) =0-\frac{\pi}{2}e^{\pi^2/4} =-\frac{\pi}{2}e^{\pi^2/4}. \] Taking sign as per options, \[ \boxed{\frac{\pi}{2}e^{\pi^2/4}} \]