Question

Let $F\left(x\right)=sinx{\int }_0^{x}costdt+2{\int }_0^{x}tdt+co{s}^{2}x-x^2$ . Then area bounded by xF(x) and ordinate x = 0 and x = 5 with x- axis is

• A. $16$
• B. $\frac{25}{2}$
• C. $\frac{35}{2}$
• D. $25$

Answer 1

The correct option is B $\frac{25}{2}$

$F\left(x\right)=sinx{\int }_0^{x}costdt+2{\int }_0^{x}tdt-x^2+co{s}^{2}x.$
$=sinx(sint{)}_0^x+2{\left(\frac{t^2}{2}\right)}_0^x-x^2+co{s}^{2}x$
$=si{n}^{2}x+x^2-x^2+co{s}^{2}x=1$
$A={\int }_0^{5}xF\left(x\right)dx={\int }_0^5\left(x\right)\left(1\right)dx={\left[\frac{x^2}{2}\right]}_0^5=\frac{25}{2}$