Question

Let $f\left(x\right)=\text{sin}x{\int }_0^x\text{cos t dt}+2{\int }_0^x\text{t dt}+{\text{cos}}^{2}x-x^2$. Then area bounded by $xf\left(x\right)$ 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

$\frac{25}{2}$

$f\left(x\right)=\text{sin}x{\int }_0^x\text{cos t dt}+2{\int }_0^x\text{t dt}-x^2+{\text{cos}}^{2}x$

$=\text{sin}x(\text{sin t}{)}_0^x+2{\left(\frac{t^2}{2}\right)}_0^x-x^2+{\text{cos}}^{2}x$

$={\text{sin}}^{2}x+x^2-x^2+{\text{cos}}^{2}x=1$

$A={\int }_0^{5}xf\left(x\right)dx={\int }_0^5\left(x\right)\left(1\right)dx={\left(\frac{x^2}{2}\right)}_0^5=\frac{25}{2}$