Answer to Question #259873 in Calculus for Dhanush

Question

use the fundamental theorem of integral calculus to evaluate the integral integration of 0 to 1 ( 2xsinx 1/x - cos.1/x )dx

Answer 1

$$\displaystyle\int_{0}^{1}(2x\sin(1/x)-\cos(1/x))dx$$

Let $t=x^2\sin(1/x).$ Then

$$dt=(2x\sin(1/x)-\cos(1/x))dx$$

$$\int(2x\sin(1/x)-\cos(1/x))dx=\int dt=t+C$$

$$=x^2\sin(1/x)+C$$

$$\displaystyle\int_{0}^{1}(2x\sin(1/x)-\cos(1/x))dx$$

$$=\lim\limits_{A\to0^+}\displaystyle\int_{A}^{1}(2x\sin(1/x)-\cos(1/x))dx$$

$$=\lim\limits_{A\to0^+}[x^2\sin(1/x)]\begin{matrix} 1 \\ A \end{matrix}$$

$$=1^2\sin(1/1)-\lim\limits_{A\to0^+}(A^2\sin(1/A))$$

Use the Squeeze Theorem

$$-A^2\leq A^2\sin(1/A)\leq A^2, A\in \R$$

$$\lim\limits_{A\to0^+}(-A^2)=\lim\limits_{A\to0^+}(A^2)=0$$

Then by the Squeeze Theorem

$$\lim\limits_{A\to0^+}(A^2\sin(1/A))=0$$

Therefore

$$\displaystyle\int_{0}^{1}(2x\sin(1/x)-\cos(1/x))dx=\sin(1)$$