Integral Calculus: answer key to the final year math exercises.

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The answer to the math exercises in the final year of high school on the calculation of an integral using an intermediate integral, as well as the property of linearity (additivity).

Exercise 1:

Calculate

$I=\,\int_{1}\;^{2}\frac{1+x^2}{1+x}dx$

by looking for an intermediate integral of the form

$J\,=\,\int_{1}\;^{2}\frac{f(x)}{g(x)}dx$ that will easily fit in.

We consider the integral:

$J=\,\int_{1}\;^2\,\frac{2x}{x+1}dx\,=\,%5B2ln(x+1)%5D_{1}^2\,=\,2ln\frac{3}{2}$

Let’s calculate:

$I\,+\,J\,=\,\int_{1}\;^2\,\frac{1+x^2}{1+x}dx\;\,+\,\int_{1}\;^2\frac{2x}{x+1}\,dx=\,\int_{1}\;^2\,\frac{1+x^2+2x}{x+1}\,dx=\,\int_{1}\;^2\frac{(x+1)^2}{x+1}dx\,=\int_{1}\;^2\,(x+1)dx\,=\,%5B\frac{x^2}{2}\,+x%5D_{1}^2\,=\frac{5}{2}$

$I\,=\,\frac{5}{2}-2ln(\frac{3}{2})$

Exercise 2:

Calculate these integrals by integrating by partials:

A. $\int_{0}^{3}x\sqrt{3-x}dx$ .

Let u=x u’=1

$v'=\sqrt{3-x}$ and $v=-\frac{2}{3}(3-x)^{\frac{^3}{2}}$

$A=%5B-\frac{2}{3}x(3-x)^{\frac{3}{2}}%5D+\int_{0}^{3}\,\frac{2}{3}(3-x)^{\frac{3}{2}}dx$

$A=0+\,\frac{2}{3}\int_{0}^{3}(3-x)^{\frac{3}{2}}dx$

$A=\frac{2}{3}%5B-\frac{2}{5}(3-x)^{\frac{5}{2}}%5D$

$A=-\frac{2}{3}\times \,\frac{2}{5}%5B(3-x)^{\frac{5}{2}}%5D$

$A=-\frac{4}{15}\times \,\,(\,-3^{\frac{5}{2}}\,\,)$

$A=\frac{4}{15}\times \,3^{\frac{5}{2}}$

$A=\frac{4}{15}\times \,3^2\times \,3^{\frac{1}{2}}$

${\color{DarkRed}\,A=\frac{12}{5}\sqrt{3}}$

Exercise 3:

Let be the function defined on $\mathbb{R}^{+*}$ by $f(x)=\frac{1}{x}+\frac{lnx}{x}$ .

What is the derivative of on $\mathbb{R}^{+*}$ ?

$f'(x)=-\frac{1}{x^2}+\frac{\frac{1}{x}\times \,x-lnx\times \,1}{x^2}$

$f'(x)=-\frac{1}{x^2}+\frac{1-lnx}{x^2}$

$f'(x)=\frac{-1+1-lnx}{x^2}$

$f'(x)=-\frac{lnx}{x^2}$

Exercise 4:

Exercise 5:

Exercise 6:

Consider the sequence $(\,U_n)$ defined, for any natural number n , by :

$U_n=\int_{0}^{1}\frac{e^{-nx}}{1+e^{-x}}dx$

a. Show that .

b. Calculate , deduce .

2. Show that, for any natural number n, $U_n\geq\,\,0$ .

a. Show that, for any non-zero natural number n,

$U_{n+1}+U_n=\frac{1-e^{-n}}{n}$

b. Deduce that, for any non-zero natural number n,

$U_n\leq\,\,\frac{1-e^{-n}}{n}$

4. Determine the limit of the sequence $(\,U_n)$ .

Exercise 7:

Exercise 8:

Exercise 9:

Calculate:

$I=\int_{1}^{x}(t^2-t)ln\,t\,dt$

Let’s ask:

$u=lnt\,\,u'=\frac{1}{t}$ and $v'=t^2-t\,\,\,\,v=\frac{t^3}{3}-\frac{t^2}{2}$

$I=,%5B,(\frac{t^3}{3}-\frac{t^2}{2})lnt,,%5D-\int_{1}^{x}\frac{1}{t}\times ,(\frac{t^3}{3}-\frac{t^2}{2})$

$I=(\frac{x^3}{3}-\frac{x^2}{2})lnx\,,-\int_{1}^{x}\frac{1}{t}\times ,(\frac{t^3}{3}-\frac{t^2}{2})$

$I=(\frac{x^3}{3}-\frac{x^2}{2})lnx\,,-\int_{1}^{x},\frac{t^2}{3}-\frac{t}{2}$

$I=(\frac{x^3}{3}-\frac{x^2}{2})lnx\,,-,%5B,\frac{t^3}{9}-\frac{t^2}{4},%5D$

$I=(\frac{x^3}{3}-\frac{x^2}{2})lnx\,,-(,\frac{x^3}{9}-\frac{x^2}{4}-\frac{1}{9}+\frac{1}{4})$

$I=(\frac{x^3}{3}-\frac{x^2}{2})lnx\,,-\frac{x^3}{9}+\frac{x^2}{4}+\frac{1}{9}-\frac{1}{4}$

$I=(\frac{x^3}{3}-\frac{x^2}{2})lnx\,,-\frac{x^3}{9}+\frac{x^2}{4}-\frac{5}{36}$

Exercise 10:

$f(x)=-1+x+\frac{x^3}{2}+x^5\,\\\,F(x)=-x+\frac{x^2}{2}+\frac{x^4}{8}+\frac{x^6}{6}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=(x-1)^2(x+1)=(x-1)(x^2-1)=x^3-x^2-x+1\,\\\,F(x)=\frac{x^4}{4}-\frac{x^3}{3}-\frac{x^2}{2}+x+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=\frac{5}{4}x+\frac{7}{3}x^2+{1}{2}x^4\,\\\,F(x)=\frac{5}{8}x^2+\frac{7}{9}x^3+\frac{12}{5}x^5+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=x+\frac{1}{\sqrt{x}}\,\\\,F(x)=\frac{x^2}{2}+2\sqrt{x}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=-\frac{1}{x^2}+\frac{1}{x^3}\,\\\,F(x)=\frac{1}{x}+\frac{-1}{2x^2}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=\frac{-3}{(3x-1)^2}\,.$

$F(x)=-\frac{1}{(3x-1)}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=cos(x)+sin(x)\,.$

$F(x)=sin(x)-sin(x)+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=\frac{-3}{\sqrt{6x+7}}\,.$

$F(x)=\frac{-\sqrt{6x+7}}{2}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=\frac{1}{cos^2(x)}+cos\,x\,.$

$F(x)=tan\,x\,+\,sinx+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=ln(x)\,.$

$F(x)=xln(x)-x+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=sinx\times \,cos^2\,x\,.$

$F(x)=-\frac{\,cos^3\,x}{3}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=sin(-3x+1)\,.$

$F(x)=\frac{cos(-3x+1)}{3}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=\frac{3x}{(x^2+1)^2}\,.$

$F(x)=-\frac{3}{2(x^2+1)}+k\,\,(k\,\in\,\mathbb{R}\,).$

$f(x)=\frac{-1}{sin^2\,x}=\frac{-cos^2\,x-sin^2\,x}{sin^2\,x}\,.$

$F(x)=\frac{cos\,x}{sin\,x}+k=\frac{1}{tanx}+k=cotan\,x+k\,\,(k\,\in\,\mathbb{R}\,).$

Exercise 12:
We consider three real numbers a,b,c such that, for all $\,x\in\,%5D0;+\infty%5B$ :

$\,\frac{1}{x(1+x)^2}=\frac{a}{x}+\frac{b}{1+x}+\frac{c}{(1+x)^2}$ .

$\,\frac{1}{x(1+x)^2}=\frac{a(1+x)^2}{x(1+x)^2}+\frac{bx(1+x)}{x(1+x)^2}+\frac{cx}{x(1+x)^2}$ .

$\,\frac{1}{x(1+x)^2}=\frac{(a+b)x^2+(2a+b+c)x+a}{x(1+x)^2}$ .

By identification, we obtain the system :

${a+b\,=\,0\,\\\,2a+b+c\,=\,0\,\\\,a\,=\,1\,$

${b\,=\,-1\,\\\,c\,=\,-1\,\\\,a\,=\,1\,$

$\,\frac{1}{x(1+x)^2}=\frac{1}{x}-\frac{1}{1+x}-\frac{1}{(1+x)^2}$

2. Let $\,X\ge\,1$ .

a. Calculate

$\,\int_{1}^{X}\,\frac{dx}{x(1+x)^2}=\int_{1}^{X}\,\frac{1}{x}-\frac{1}{1+x}-\frac{1}{(1+x)^2}dx$ .

$\,=%5Bln(x)-ln(1+x)+\frac{1}{1+x}%5D$

$\,=ln(X)-ln(1+X)+\frac{1}{1+X}+ln\,2\,-\frac{1}{2}$

$\,=ln(\frac{X}{1+X})+\frac{1}{1+X}+ln\,2\,-\frac{1}{2}$

b. Let f be the function defined on $\,x\in\,%5B1;+\infty%5B$ by $\,f(X)=\int_{1}^{X}\,\frac{ln\,x}{(1+x)^3}dx$

By integrating by parts, calculate f(X) as a function of X .

Let’s ask:

u(x)= lnx then $\,u'(x)=\frac{1}{x}$

$\,v'(x)=\frac{1}{(1+x)^3$ then $v(x)=\,\frac{-1}{2(1+x)^2}$

So:

$\,f(X)=\int_{1}^{X}\,\frac{ln\,x}{(1+x)^3}dx=%5B\frac{-lnx}{2(1+x)^2}%5D+\int_{1}^{X}\,\frac{1}{2x(1+x)^2}dx$

$\,f(X)=\int_{1}^{X}\,\frac{ln\,x}{(1+x)^3}dx=\frac{-lnX}{2(1+X)^2}+\frac{1}{2}\int_{1}^{X}\,\frac{1}{x(1+x)^2}dx$

$\,f(X)=\int_{1}^{X}\,\frac{ln\,x}{(1+x)^3}dx=\frac{-lnX}{2(1+X)^2}+\frac{1}{2}(ln(\frac{X}{1+X})+\frac{1}{1+X}+ln\,2\,-\frac{1}{2})$

$\,f(X)=\int_{1}^{X}\,\frac{ln\,x}{(1+x)^3}dx=\frac{-lnX}{2(1+X)^2}+\frac{1}{2}(ln(\frac{X}{1+X})+\frac{1}{1+X})+\frac{1}{2}(ln\,2\,-\frac{1}{2})$

c. gold :

$\,\lim_{X\,\to\,+\infty}\,\frac{-lnX}{2(1+X)^2}+\frac{1}{2}(ln(\frac{X}{1+X})+\frac{1}{1+X})=0$

therefore:

$\,\lim_{X\,\to\,+\infty}\,f(X)=\frac{1}{2}(ln2-\frac{1}{2})$

Exercise 14:
Calculate the proposed integral:

b. $\int_{-1}^{2}-x+6dx=%5B-\frac{x^2}{2}+6x%5D=-\frac{2^2}{2}+12+\frac{1}{2}+6=16,5$

c. $\int_0^{4}\,(2x^2+8x-1)\,dx=%5B2\frac{x^3}{3}+4x^2-x%5D\\=2\times \frac{4^3}{3}+4\times \,4^2-4\,-2\times \frac{0^3}{3}-4\times \,0^2+0\\=\frac{128}{3}+64-4=\frac{308}{3}.$

d. $\int_0^{\frac{2\pi}{3}}\,(cosx)\,dx=%5Bsin\,x%5D\\=\sin{\,\frac{2\pi}{3}}\,-sin\,0\,=-\frac{\sqrt{3}}{2}.$

e. $\int_{-2}^{0}\,(x^5+4x^3+x^2-x)\,dx\,=%5B\frac{x^6}{6}+x^4+\frac{x^3}{3}-\frac{x^2}{2}%5D\\=\frac{0^6}{6}+0^4+\frac{0^3}{3}-\frac{0^2}{2}\,-\frac{(-2)^6}{6}-(-2)^4-\frac{(-2)^3}{3}+\frac{(-2)^2}{2}\,\\=-\frac{64}{6}-16-\frac{8}{3}+2=-\frac{66}{3}.$

f. $\int_1^{3}\,(\frac{1}{x^2})\,dx\,\\=\,%5B-\frac{1}{x}%5D=-\frac{1}{3}+\frac{1}{1}=\frac{2}{3}.$

g. $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}(\frac{1}{cos^2\,x})\,dx\\=2\times \,\int_{0}^{\frac{\pi}{4}}(\frac{1}{cos^2\,x})\,dx=2\times \,%5Bln(cos\,x)%5D=2ln(\frac{\sqrt{2}}{2}).$

h. $\int_{3}^{4}(\frac{1}{\sqrt{2x+5}})\,dx=%5B\sqrt{2x+5}%5D\\=\sqrt{2\times \,4+5}-\sqrt{2\times \,3+5}\\=\sqrt{13}-\sqrt{11}.$

Exercise 15:

Either $f(x)=x^2+1\,\,.$

$\int_{-1}^{0}(x^2+1)\,dx=%5B\frac{x^3}{3}+x%5D=\frac{0^3}{3}+0-\frac{(-1)^3}{3}+1=\frac{2}{3}.$

Exercise 16:

We consider

$\int_a^{b}\,f(x)\,dx=5\,\,.$ and $\int_a^{b}\,g(x)\,dx=3\,\,.$

a. $\int_a^{b}\,(2f(x)-4g(x))\,dx=-2\,\,.$

b. $\beta\,=6.$

Exercise 17:

$\int_{-5}^{5}\,(x^3-tan\,x)\,dx=0\,\,.$

because the function is odd on the interval [-5;5] centered in 0.

Answers to the exercises on integrals in high school.

After having consulted the answers to these exercises on integral calculus in the final year of high school, you can return to the exercises in the final year of high school.

Senior year exercises.

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