Answer to Question #228471 in Calculus for Savelile

Question

A)Determine the following integrals:

(i) $\int$ (x-2/x^2)(X+2/x^2)dx

(ii) $\int$ e^5x(e^2x/7+3/e^3x)dx

(iii) $\int$ 1/(4-√3x)^3 dx

(iv) $\int$ ^π/4₀(tan x)^3(sec x)^3 dx

B)let f(x)=x^2-2 and g(x)=-|x|,then

(i) sketch the graphs of f and g on the same axes.

(ii) find the area enclosed by f(x)=x^2-2 and g(x)=-|x|.

Answer 1

$A\\ i)\\ \int [x-\frac{2}{x^2}][x+\frac{2}{x^2}]dx\\ =\int [x^2-\frac{4}{x^4}]dx\\ = \frac{x^3}{3}+\frac{4}{3x^3}+c\\ ii)\\ \int e^{5x}[\frac{e^{2x}}{7}+\frac{3}{e^{3x}}]dx\\ =\int [\frac{e^{7x}}{7}+3{e^{2x}}]dx\\ =\frac{e^{7x}}{49}+\frac{3e^{2x}}{2}+c\\ iii)\\ \int \frac{1}{(4-\sqrt{3x})^3}dx\\ \text{put}\\ 4-\sqrt{3x}=t\\ =>\frac{-3}{2\sqrt{3x}}dx=dt\\ \implies dx=\frac{-2}{3}(4-t)dt\\ \int \frac{1}{(4-\sqrt{3x})^3}dx\\ =\frac{-2}{3} \int \frac{4-t}{t^3}dt\\ =\frac{-2}{3} \int (4t^{-3}-t^{-2})dt\\ =\frac{-2}{3}(\frac{-2}{t^{2}}+\frac{1}{3t^3})+c\\ =\frac{-2}{3}(\frac{-2}{(4-\sqrt{3x}) ^{2}}+\frac{1}{3(4-\sqrt{3x})^3})+c\\ iv)\\ \int^{π/4}_{0 } (tan x)^3 (sec x )^3dx\\ \text{put},\\ secx=t\\ =>tanx \space secx \space dx=dt\\ \text{and using }tan^2x=1-sec^2x, \text{we get}\\ \int^{π/4}_{0 } (tan x)^3 (sec x )^3dx\\ =\int^{π/4}_{0 } (secx)^2(tan x )^2tanx \space secx\space dx\\ =\int^{π/4}_{0} (sec x )^2(1-sec^2x)tanx \space secx\space dx\\ =\int^{\sqrt2}_1 t^2(1-t^2)dt\\ =\int^{\sqrt2}_1 (t^2-t^4)dt\\ =[\frac{t^3}{3}-\frac{t^5}{5}]^{\sqrt2}_1\\ =-0.32\\ B\\ i)$

blue curve denotes $f=x^2-2$

red curve denotes g=-|x|

(ii)

area of region I i.e., ABC,

$2×\frac{1}{2}×1×1\\ =1$

area of region II i.e., ADC,

$2×\int_{-1}^1 -(x^2-2)dx\\ =2×\int_{-1}^1 (2-x^2)dx\\ =2[2x-\frac{x^3}{3}]_{-1}^1\\ =2[2-\frac{1}{3}+2-\frac{1}{3}]\\ =\frac{20}{3}$

total area of the region bounded i.e., ABCD is 23/3.