Mathematics - Definite Integration Question with Solution | TestHub

(A) $I={\int }_4^{10}\frac{\left[x^2\right]dx}{\left[x^2-28x+196\right]+\left[x^2\right]}=$

$$\begin{array}{rl}& I={\int }_4^{10}\frac{\left[x^2-28x+196\right]dx}{\left[x^2\right]+\left[x^2-28x+196\right]}\\ & 2I={\int }_4^{10}dx\\ & I=3\end{array}$$

(B) $I={\int }_{-1}^2\frac{|x|dx}{x}$

$$\begin{array}{rl}& I=-{\int }_{-1}^{0}dx+{\int }_0^{2}dx\\ & I=-[0-(-1)]+(2-0)\\ & I=-1+2\\ & I=1\end{array}$$

(C) ${\lim }_{x\to \infty }\frac{1^{99}+2^{99}+3^{99}\dots \dots x^{99}}{x^{100}}=2\mathrm{A}$

$$\begin{array}{rl}& 2\mathrm{A}=\underset{\mathrm{x}\to \infty }{\lim }\frac{1}{\mathrm{x}}\frac{{\sum }_{\mathrm{r}=1}^{\mathrm{x}}{\mathrm{r}}^{99}}{{\mathrm{x}}^{99}}\\ & 2\mathrm{A}=\underset{\mathrm{x}\to \infty }{\lim }\frac{1}{\mathrm{x}}\sum _{\mathrm{r}=1}^{\mathrm{x}}{\left(\frac{\mathrm{r}}{\mathrm{x}}\right)}^{99}\\ & 2\mathrm{A}={\int }_0^1{\mathrm{x}}^{99}\mathrm{dx}\\ & 2\mathrm{A}={\left[\frac{{\mathrm{x}}^{100}}{100}\right]}_0^1\\ & \mathrm{A}=\frac{1}{200}\text{ then value of }\mathrm{A}\end{array}$$

(D) $I=5050{\int }_{-1}^1\left|x^{200}\right|\mathrm{dx}$

$$\begin{array}{rl}& I=5050\times 2{\int }_0^1{x}^{200}dx\\ & \frac{1}{\alpha }=5050\times 2\times {\left[\frac{x^{201}}{201}\right]}_0^1\\ & \frac{1}{\alpha }=10100\times \frac{1}{201}\\ & \alpha =\frac{201}{10100}\end{array}$$