Short Question 2 – 4

Posted on April 22, 2020 by user

Question 2:

Given that

$Given that \int \frac{4}{{(1 + x)}^{4}} d x = m {(1 + x)}^{n} + c,$

find the values of m and n.

Solution:

$\begin{array}{l} \int \frac{4}{{(1 + x)}^{4}} d x = m {(1 + x)}^{n} + c \\ \int 4 {(1 + x)}^{- 4} d x = m {(1 + x)}^{n} + c \\ \frac{4 {(1 + x)}^{- 3}}{- 3 (1)} + c = m {(1 + x)}^{n} + c \\ - \frac{4}{3} {(1 + x)}^{- 3} + c = m {(1 + x)}^{n} + c \\ m = - \frac{4}{3}, n = - 3 \end{array}$

Question 3:

$\begin{array}{l} Given \int_{- 1}^{2} 2 g (x) d x = 4, and \int_{- 1}^{2} [m x + 3 g (x)] d x = 15. \\ Find the value of constant m . \end{array}$

Solution:

$\begin{array}{l} \int_{- 1}^{2} [m x + 3 g (x)] d x = 15 \\ \int_{- 1}^{2} m x d x + \int_{- 1}^{2} 3 g (x) d x = 15 \\ {[\frac{m x^{2}}{2}]}_{- 1}^{2} + 3 \int_{- 1}^{2} g (x) d x = 15 \\ [\frac{m {(2)}^{2}}{2} - \frac{m {(- 1)}^{2}}{2}] + \frac{3}{2} \int_{- 1}^{2} 2 g (x) d x = 15 \\ 2 m - \frac{1}{2} m + \frac{3}{2} (4) = 15 \leftarrow given \int_{- 1}^{2} 2 g (x) d x = 4 \\ \frac{3}{2} m + 6 = 15 \\ \frac{3}{2} m = 9 \\ m = 9 \times \frac{2}{3} \\ m = 6 \end{array}$

Question 4:

$Given \frac{d}{d x} (\frac{2 x}{3 - x}) = g (x), find \int_{1}^{2} g (x) d x .$

Solution:

$\begin{array}{l} Given \frac{d}{d x} (\frac{2 x}{3 - x}) = g (x) \\ \int g (x) d x = \frac{2 x}{3 - x} \\ Thus, \\ \int_{1}^{2} g (x) d x = {[\frac{2 x}{3 - x}]}_{1}^{2} \\ = \frac{2 (2)}{3 - 2} - \frac{2 (1)}{3 - 1} \\ = 4 - 1 \\ = 3 \end{array}$