curriculum-current
Area between two curves and ∫ 0 1 ln x d x \int_0^1\ln x\,\mathrm{d}x ∫ 0 1 ln x d x
Math I: Calculus · Mock exam, Question 5(b)–(c) · 14 points · Lectures 11–12
Problem
(b) Find the area enclosed by 2 y 2 − y = x + 2 2y^2-y=x+2 2 y 2 − y = x + 2 y 2 = x . y^2=x\text{.} y 2 = x .
(c) Evaluate ∫ 0 1 ln x d x \int_0^1\ln x\,\mathrm{d}x ∫ 0 1 ln x d x
Plan
In (b) both curves are easy to write as x x x y , y\text{,} y , y : y\text{:} y : = ∫ ( right − left ) d y . =\int(\text{right}-\text{left})\,\mathrm{d}y\text{.} = ∫ ( right − left ) d y . ln x \ln x ln x 0 , 0\text{,} 0 , x = e − t , x=\mathrm{e}^{-t}\text{,} x = e − t , [ 0 , ∞ ) . [0,\infty)\text{.} [ 0 , ∞ ) .
Step-by-step solution
Try each step yourself before you open it.
(b) Step 1 · Write both curves as functions of y (1 point)
x = 2 y 2 − y − 2 , x = y 2 . x=2y^2-y-2,\quad x=y^2. x = 2 y 2 − y − 2 , x = y 2 .
(b) Step 2 · Find the intersections (2 points)
y 2 = 2 y 2 − y − 2 y^2=2y^2-y-2 y 2 = 2 y 2 − y − 2 y 2 − y − 2 = 0 , y^2-y-2=0\text{,} y 2 − y − 2 = 0 , ( y − 2 ) ( y + 1 ) = 0 . (y-2)(y+1)=0\text{.} ( y − 2 ) ( y + 1 ) = 0 . y = − 1 y=-1 y = − 1 y = 2 . y=2\text{.} y = 2 .
(b) Step 3 · Which curve is on the right? (1 point)
For − 1 < y < 2 , -1<y<2\text{,} − 1 < y < 2 ,
y 2 − ( 2 y 2 − y − 2 ) = − ( y − 2 ) ( y + 1 ) > 0 , y^2-(2y^2-y-2)=-(y-2)(y+1)>0, y 2 − ( 2 y 2 − y − 2 ) = − ( y − 2 ) ( y + 1 ) > 0 , so x = y 2 x=y^2 x = y 2 x = 2 y 2 − y − 2 x=2y^2-y-2 x = 2 y 2 − y − 2
(b) Step 4 · Set up and evaluate (2 points)
Area = ∫ − 1 2 ( − y 2 + y + 2 ) d y = [ − y 3 3 + y 2 2 + 2 y ] − 1 2 = 10 3 − ( − 7 6 ) = 9 2 . \begin{aligned}
\text{Area}&=\int_{-1}^{2}\left(-y^2+y+2\right)\,\mathrm{d}y\\
&=\left[-\frac{y^3}{3}+\frac{y^2}{2}+2y\right]_{-1}^{2}\\
&=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{9}{2}.
\end{aligned} Area = ∫ − 1 2 ( − y 2 + y + 2 ) d y = [ − 3 y 3 + 2 y 2 + 2 y ] − 1 2 = 3 10 − ( − 6 7 ) = 2 9 . Concept: Fundamental theorem of calculus, part 2
(c) Method 1, Step 1 · Write the improper integral as a limit (1 point)
ln x → − ∞ \ln x\to-\infty ln x → − ∞ x → 0 + , x\to0^+\text{,} x → 0 + ,
∫ 0 1 ln x d x = lim ε → 0 + ∫ ε 1 ln x d x . \int_0^1\ln x\,\mathrm{d}x=\lim_{\varepsilon\to0^+}\int_{\varepsilon}^{1}\ln x\,\mathrm{d}x. ∫ 0 1 ln x d x = ε → 0 + lim ∫ ε 1 ln x d x . Concept: Improper integral
(c) Method 1, Step 2 · Integrate by parts (1 point)
With u = ln x u=\ln x u = ln x v ′ = 1 , v'=1\text{,} v ′ = 1 ,
∫ ε 1 ln x d x = [ x ln x ] ε 1 − ∫ ε 1 1 d x = − ε ln ε − ( 1 − ε ) . \int_{\varepsilon}^{1}\ln x\,\mathrm{d}x =\big[x\ln x\big]_{\varepsilon}^{1}-\int_{\varepsilon}^{1}1\,\mathrm{d}x =-\varepsilon\ln\varepsilon-(1-\varepsilon). ∫ ε 1 ln x d x = [ x ln x ] ε 1 − ∫ ε 1 1 d x = − ε ln ε − ( 1 − ε ) .
(c) Method 1, Step 3 · Take the limit (2 points)
ε ln ε → 0 \varepsilon\ln\varepsilon\to0 ε ln ε → 0 ε → 0 + . \varepsilon\to0^+\text{.} ε → 0 + .
lim ε → 0 + ln ε 1 / ε = lim ε → 0 + 1 / ε − 1 / ε 2 = lim ε → 0 + ( − ε ) = 0. \lim_{\varepsilon\to0^+}\frac{\ln\varepsilon}{1/\varepsilon} =\lim_{\varepsilon\to0^+}\frac{1/\varepsilon}{-1/\varepsilon^2} =\lim_{\varepsilon\to0^+}(-\varepsilon)=0. ε → 0 + lim 1/ ε ln ε = ε → 0 + lim − 1/ ε 2 1/ ε = ε → 0 + lim ( − ε ) = 0. So the integral equals 0 − 1 = − 1 . 0-1=-1\text{.} 0 − 1 = − 1 .
(c) Method 2, Step 1 · Substitute x = e^(−t) (2 points)
Let x = e − t . x=\mathrm{e}^{-t}\text{.} x = e − t . ln x = − t \ln x=-t ln x = − t d x = − e − t d t ; \mathrm{d}x=-\mathrm{e}^{-t}\,\mathrm{d}t\text{;} d x = − e − t d t ; x = 1 x=1 x = 1 t = 0 t=0 t = 0 x → 0 + x\to0^+ x → 0 + t → ∞ . t\to\infty\text{.} t → ∞ .
∫ 0 1 ln x d x = ∫ ∞ 0 ( − t ) ( − e − t ) d t = − ∫ 0 ∞ t e − t d t . \int_0^1\ln x\,\mathrm{d}x =\int_{\infty}^{0}(-t)\left(-\mathrm{e}^{-t}\right)\,\mathrm{d}t =-\int_0^{\infty}t\mathrm{e}^{-t}\,\mathrm{d}t. ∫ 0 1 ln x d x = ∫ ∞ 0 ( − t ) ( − e − t ) d t = − ∫ 0 ∞ t e − t d t . The orientation flips because x x x t t t
Concept: Integration by substitution, type 2
(c) Method 2, Step 2 · Evaluate the new integral (2 points)
On [ 0 , R ] , [0,R]\text{,} [ 0 , R ] ,
∫ 0 R t e − t d t = [ − t e − t ] 0 R + ∫ 0 R e − t d t = − R e − R + 1 − e − R → 1 , \int_0^R t\mathrm{e}^{-t}\,\mathrm{d}t =\big[-t\mathrm{e}^{-t}\big]_0^R+\int_0^R\mathrm{e}^{-t}\,\mathrm{d}t =-R\mathrm{e}^{-R}+1-\mathrm{e}^{-R}\to1, ∫ 0 R t e − t d t = [ − t e − t ] 0 R + ∫ 0 R e − t d t = − R e − R + 1 − e − R → 1 , because R e − R → 0 R\mathrm{e}^{-R}\to0 R e − R → 0 e − R → 0 . \mathrm{e}^{-R}\to0\text{.} e − R → 0 . ∫ 0 1 ln x d x = − 1 . \int_0^1\ln x\,\mathrm{d}x=-1\text{.} ∫ 0 1 ln x d x = − 1 .
Where marks are lost
Integrating in x x x It can be done, but the region then needs several pieces. Integrating in y y y Treating ∫ 0 1 ln x d x \int_0^1\ln x\,\mathrm{d}x ∫ 0 1 ln x d x Write the limit; the mark for justifying ε ln ε → 0 \varepsilon\ln\varepsilon\to0 ε ln ε → 0 Losing the sign in the substitution. Swapping the limits and the minus sign from d x \mathrm{d}x d x Giving the same method twice. "Two different ways" means two genuinely different routes.
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Symbols: epsilon, infinity, integral, phi, sum