Short Question 8 – 10

Question 8:

Given y = \frac{5 x}{x^{2} + 1} and \frac{d y}{d x} = g (x), find the value of \int_{0}^{3} 2 g (x) d x .

Solution:

\begin{array}{l} Since \frac{d y}{d x} = g (x), thus y = \int g (x) d x \\ \int_{0}^{3} 2 g (x) d x = 2 \int_{0}^{3} g (x) d x \\ = 2 {[y]}_{0}^{3} \\ = 2 {[\frac{5 x}{x^{2} + 1}]}_{0}^{3} \\ = 2 [\frac{5 (3)}{3^{2} + 1} - 0] \\ = 2 (\frac{15}{10}) \\ = 3 \end{array}

Question 9:

Find \int_{5}^{k} (x + 1) d x, in terms of k .

Solution:

\begin{array}{l} {\int_{5}^{k} (x + 1) d x = [\frac{x^{2}}{2} + x]}_{5}^{k} \\ = (\frac{k^{2}}{2} + k) - (\frac{5^{2}}{2} + 5) \\ = \frac{k^{2} + 2 k}{2} - \frac{35}{2} \\ = \frac{k^{2} + 2 k - 35}{2} \end{array}

Question 10:

\begin{array}{l} Given that = \int_{2}^{5} g (x) d x = - 2 . Find \\ (a) the value of \int_{5}^{2} g (x) d x, \\ (b) the value of m if \int_{2}^{5} [g (x) + m (x)] d x = 19 \end{array}

Solution:

\begin{array}{l} (a) \int_{5}^{2} g (x) d x = - \int_{2}^{5} g (x) d x \\ = - (- 2) \\ = 2 \end{array}

\begin{array}{l} (b) \int_{2}^{5} [g (x) + m (x)] d x = 19 \\ \int_{2}^{5} g (x) d x + m \int_{2}^{5} x d x = 19 \\ - 2 + m {[\frac{x^{2}}{2}]}_{2}^{5} = 19 \\ \frac{m}{2} {[x^{2}]}_{2}^{5} = 21 \\ \frac{m}{2} [25 - 4] = 21 \\ 21 m = 42 \\ m = 2 \end{array}