Tutor profile: Abhijith C.

Three points $$ P(2,1) $$, $$ Q(5,-3) $$, $$ R(2,-7) $$ form the vertices of a triangle. Identify which type of a triangle it is?

Concept: From the given points, we can easily calculate the length of each of the three sides of the triangle. And the relative length of these three sides will help us determine which type of a triangle it is. Distance between two points $$ A(x_1,y_1)$$ and $$ B(x_2,y_2)$$ in Cartesian Coordinate system is given by the formula: $$ AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Calculation: Given vertices are $$ P (2,1) $$, $$ Q(5,-3) $$ and $$ R(2,-7) $$ Distance between P and Q, $$ PQ = \sqrt{(5-2)^2+(-3-1)^2} = \sqrt{3^2+(-4)^2}= \sqrt{9+16}= \sqrt{25}=5$$ Distance between Q and R, $$ QR = \sqrt{(2-5)^2+(-7+3)^2} = \sqrt{(-3)^2+(-4)^2}= \sqrt{9+16}= \sqrt{25}=5$$ Distance between P and R, $$ PR = \sqrt{(2-2)^2+(-7-1)^2} = \sqrt{(-8)^2}= \sqrt{64}=8 $$ PQ and QR are of the same length. But PR has a different length. Since two sides of this triangle are equal in length, the given points are vertices of an Isosceles triangle.

For the data $$ x : $$ 0 1 2 3 $$ f(x) : $$ 2 4 3 5, find the value of $$ \int_0^3 {[f(x)]}^3 \mathrm{d}x $$ by applying Trapezoidal Rule.

Concept: Trapezoidal Rule is given by $$ \int_a^b {f(x)} \mathrm{d}x = \frac{h}{2}[(y_0+y_n)+ 2(y_1+y_2+y_3+...y_{n-1})]$$ where, b is the upper limit a is the lower limit n is the number of intervals h is the step size and is given by the formula $$ h = \frac{b-a}{n} $$ Calculation: Given data is: $$ x : $$ 0 1 2 3 $$ f(x) : $$ 2 4 3 5 , Here, $$ y_n = [f(x)]^3 $$ $$ y_n : $$ $$ y_0 = 8 ; $$ $$ y_1 = 64 ; $$ $$ y_2 = 27 ; $$ $$ y_3 = 125 $$ upper limit, $$ b = 3 $$ lower limit, $$ a = 0 $$ number of intervals, $$ n = 3 $$ $$ \therefore $$ step size, $$h = \frac{b-a}{n} = \frac{3-0}{3} = 1 $$ So by using Trapezoidal Rule, we get $$ \int_0^3 {[f(x)]}^3 \mathrm{d}x = \frac{h}{2}[(y_0+y_3)+ 2(y_1+y_2)]$$ $$ = \frac{1}{2}[(8+125)+ 2(64+27)]$$ $$ = \frac{315}{2}$$ $$ = 157.5$$

Evaluate $$ \int \frac{{\mathrm{e}}^{x}}{\mathrm{e}^{2x}-9} \mathrm{d}x $$ .

Concept: Given integral can be evaluated easily once it is reduced to the form of a Standard Integral as given below: $$ \int\frac{1}{x^2-a^2} \mathrm{d}x = \frac{1}{2a}\log{|\frac{x-a}{x+a}|} + C , x > a $$ Calculation: Let $$ I= \int \frac{{\mathrm{e}}^{x}}{\mathrm{e}^{2x}-9} \mathrm{d}x $$ $$ = \int \frac{{\mathrm{e}}^{x}}{({\mathrm{e}^{x}})^2-3^2} \mathrm{d}x $$ In the above equation, put $$ t= \mathrm{e}^x$$. Now, differentiating both sides w.r.t $$ x $$, we get $$ \frac{\mathrm{d}t}{\mathrm{d}x} = \mathrm{e}^x \implies \mathrm{d}t = \mathrm{e}^x \mathrm{d}x $$ $$ \therefore I= \int \frac{1}{t^2-3^2} \mathrm{d}t $$ Now it reduces to the form of the Standard Integral just mentioned above. So, we have, $$ I= \int \frac{1}{t^2-3^2} \mathrm{d}t = \frac{1}{2\times3} \log{|\frac{t-3}{t+3}|} + C $$ Putting $$ t= \mathrm{e}^x$$ back in the above equation, we get the final Answer as: $$ I= \int \frac{{\mathrm{e}}^{x}}{\mathrm{e}^{2x}-9} \mathrm{d}x $$ $$ = \int \frac{1}{t^2-3^2} \mathrm{d}t $$ $$ = \frac{1}{6} \log{|\frac{\mathrm{e}^x-3}{\mathrm{e}^x+3}|} + C$$ .