Definite Integrals and Mixed Practice — Worked Solutions

Ques­tion 1: Three direct def­i­nite inte­grals#

The prob­lem#

Eval­u­ate: (a) $\displaystyle \int_0^2(x^3 + 2x)\,dx$ (b) $\displaystyle \int_1^e\frac{dx}{x}$ (c) $\displaystyle \int_0^{\pi/3}\cos x\,dx$

Under­stand­ing the prob­lem#

Each inte­gral has a stan­dard anti­deriv­a­tive. You must find it and then use the lim­its to get a num­ber.

The idea#

Use the fun­da­men­tal the­o­rem: if $F' = f$, then $\displaystyle \int_a^b f(x)\,dx = F(b) - F(a)$.

Step-by-step solu­tion#

Part (a)

Step 1. Anti­deriv­a­tive: $\displaystyle \int x^3\,dx = \frac{x^4}{4}$ and $\displaystyle \int 2x\,dx = x^2$.

$$\displaystyle \int_0^2(x^3 + 2x)\,dx = \left[\frac{x^4}{4} + x^2\right]_0^2$$

Step 2. Sub­sti­tute the lim­its.

$$\displaystyle = \left(\frac{16}{4} + 4\right) - (0 + 0) = 4 + 4 = 8$$

Part (b)

Step 1. Anti­deriv­a­tive of $\displaystyle \frac{1}{x}$ is $\log\lvert x\rvert$; on $[1, e]$, $x > 0$.

$$\displaystyle \int_1^e\frac{dx}{x} = \big[\log x\big]_1^e = \log e - \log 1 = 1 - 0 = 1$$

Part (c)

Step 1. Anti­deriv­a­tive of $\cos x$ is $\sin x$.

$$\displaystyle \int_0^{\pi/3}\cos x\,dx = \big[\sin x\big]_0^{\pi/3} = \sin\frac{\pi}{3} - \sin 0 = \frac{\sqrt{3}}{2}$$

Check­ing the answer#

In (a), the inte­grand lies between $0$ and $12$ on an inter­val of length $2$, so the answer must be between $0$ and $24$; $8$ is sen­si­ble. In (c), $\cos x$ is between $\displaystyle \tfrac{1}{2}$ and $1$ on an inter­val of length about $1.05$, and $\displaystyle \tfrac{\sqrt{3}}{2} \approx 0.87$ fits.

Answer#

(a) $8$; (b) $1$; (c) $\displaystyle \frac{\sqrt{3}}{2}$.

Ques­tion 2: Def­i­nite inte­grals by sub­sti­tu­tion#

The prob­lem#

Eval­u­ate: (a) $\displaystyle \int_0^1\frac{x}{x^2 + 1}\,dx$ (b) $\displaystyle \int_1^2\frac{dx}{x(1 + \log x)^2}$

Under­stand­ing the prob­lem#

In each inte­gral, one part is (a mul­ti­ple of) the deriv­a­tive of another part: $x$ is half the deriv­a­tive of $x^2 + 1$, and $\displaystyle \frac{1}{x}$ is the deriv­a­tive of $1 + \log x$. That sig­nals a sub­sti­tu­tion.

The idea#

Sub­sti­tute $t$ for the inner expres­sion, rewrite $dx$ in terms of $dt$, and change the lim­its to $t$-val­ues, as in Exam­ple 3. Then you never need to go back to $x$.

Step-by-step solu­tion#

Part (a)

Step 1. Let $t = x^2 + 1$. Then $dt = 2x\,dx$, so $\displaystyle x\,dx = \frac{dt}{2}$.

Step 2. Change the lim­its: $x = 0 \Rightarrow t = 1$; $x = 1 \Rightarrow t = 2$.

Step 3. Rewrite and inte­grate.

$$\displaystyle \int_0^1\frac{x}{x^2 + 1}\,dx = \frac{1}{2}\int_1^2\frac{dt}{t} = \frac{1}{2}\big[\log t\big]_1^2 = \frac{1}{2}(\log 2 - 0) = \frac{1}{2}\log 2$$

Part (b)

Step 1. Let $t = 1 + \log x$. Then $\displaystyle dt = \frac{1}{x}\,dx$.

Step 2. Change the lim­its: $x = 1 \Rightarrow t = 1 + 0 = 1$; $x = 2 \Rightarrow t = 1 + \log 2$.

Step 3. Rewrite: $\displaystyle \frac{dx}{x(1 + \log x)^2} = \frac{dt}{t^2}$.

$$\displaystyle \int_1^{1 + \log 2}t^{-2}\,dt = \left[-\frac{1}{t}\right]_1^{1 + \log 2}$$

Step 4. Sub­sti­tute the lim­its.

$$\displaystyle = -\frac{1}{1 + \log 2} - \left(-\frac{1}{1}\right) = 1 - \frac{1}{1 + \log 2}$$

Step 5. This can also be writ­ten as a sin­gle frac­tion.

$$\displaystyle 1 - \frac{1}{1 + \log 2} = \frac{\log 2}{1 + \log 2}$$

Check­ing the answer#

(a) $\displaystyle \tfrac{1}{2}\log 2 \approx 0.347$; the inte­grand rises from $0$ to $0.5$ over $[0, 1]$, so an area of about $0.35$ is rea­son­able. (b) $\displaystyle \frac{0.693}{1.693} \approx 0.409$; the inte­grand falls from $1$ to about $0.17$ on $[1, 2]$, so an area of about $0.4$ fits.

Answer#

(a) $\displaystyle \frac{1}{2}\log 2$; (b) $\displaystyle 1 - \frac{1}{1 + \log 2} = \frac{\log 2}{1 + \log 2}$.

Com­mon mis­take to avoid#

Sub­sti­tut­ing and then using the old $x$-lim­its with the new $t$-inte­gral. Change the lim­its when­ever you change the vari­able.

Ques­tion 3: A trigono­met­ric iden­tity and inte­gra­tion by parts#

The prob­lem#

Eval­u­ate: (a) $\displaystyle \int_0^{\pi/2}\cos^2 x\,dx$ (b) $\displaystyle \int_0^{1}xe^x\,dx$

Under­stand­ing the prob­lem#

(a) $\cos^2 x$ has no direct anti­deriv­a­tive in the table, so you first rewrite it. (b) is a prod­uct of $x$ and $e^x$, which calls for inte­gra­tion by parts.

The idea#

(a) Use $\displaystyle \cos^2 x = \frac{1 + \cos 2x}{2}$. (b) Use $\displaystyle \int u\,dv = uv - \int v\,du$ with $u = x$ (it becomes sim­pler when dif­fer­en­ti­ated) and $dv = e^x\,dx$.

Step-by-step solu­tion#

Part (a)

Step 1. Rewrite the inte­grand.

$$\displaystyle \int_0^{\pi/2}\cos^2 x\,dx = \int_0^{\pi/2}\frac{1 + \cos 2x}{2}\,dx$$

Step 2. Inte­grate: $\displaystyle \int\cos 2x\,dx = \frac{\sin 2x}{2}$.

$$\displaystyle = \frac{1}{2}\left[x + \frac{\sin 2x}{2}\right]_0^{\pi/2}$$

Step 3. Sub­sti­tute: $\sin\pi = 0$ and $\sin 0 = 0$.

$$\displaystyle = \frac{1}{2}\left(\frac{\pi}{2} + 0\right) - 0 = \frac{\pi}{4}$$

Part (b)

Step 1. Choose $u = x$, $dv = e^x\,dx$; then $du = dx$, $v = e^x$.

Step 2. Apply inte­gra­tion by parts with lim­its.

$$\displaystyle \int_0^1 xe^x\,dx = \big[xe^x\big]_0^1 - \int_0^1 e^x\,dx$$

Step 3. Eval­u­ate each piece.

$$\displaystyle \big[xe^x\big]_0^1 = e - 0 = e, \qquad \int_0^1 e^x\,dx = e - 1$$

Step 4. Sub­tract.

$$\displaystyle \int_0^1 xe^x\,dx = e - (e - 1) = 1$$

Check­ing the answer#

(a) By sym­me­try, $\displaystyle \int_0^{\pi/2}\cos^2 x\,dx = \int_0^{\pi/2}\sin^2 x\,dx$, and their sum is $\displaystyle \int_0^{\pi/2}1\,dx = \tfrac{\pi}{2}$; so each is $\displaystyle \tfrac{\pi}{4}$. (b) Dif­fer­en­ti­ate $e^x(x - 1)$: you get $xe^x$, and $e^x(x - 1)$ goes from $-1$ to $0$, a change of $1$.

Answer#

(a) $\displaystyle \frac{\pi}{4}$; (b) $1$.

Ques­tion 4: The inte­gral of an absolute value#

The prob­lem#

Eval­u­ate $\displaystyle \int_{-3}^{3}\lvert x - 1\rvert\,dx$.

Under­stand­ing the prob­lem#

$\lvert x - 1\rvert$ equals $x - 1$ when $x \ge 1$ and $1 - x$ when $x < 1$. So the for­mula for the inte­grand changes at $x = 1$, which lies inside $[-3, 3]$.

The idea#

Split the inter­val at $x = 1$ using prop­erty 2, $\displaystyle \int_a^b = \int_a^c + \int_c^b$, and use the cor­rect for­mula on each piece, as in Exam­ple 5.

Step-by-step solu­tion#

Step 1. Split at $x = 1$.

$$\displaystyle \int_{-3}^{3}\lvert x - 1\rvert\,dx = \int_{-3}^{1}(1 - x)\,dx + \int_{1}^{3}(x - 1)\,dx$$

Step 2. First piece.

$$\displaystyle \int_{-3}^{1}(1 - x)\,dx = \left[x - \frac{x^2}{2}\right]_{-3}^{1} = \left(1 - \frac{1}{2}\right) - \left(-3 - \frac{9}{2}\right) = \frac{1}{2} + \frac{15}{2} = 8$$

Step 3. Sec­ond piece.

$$\displaystyle \int_{1}^{3}(x - 1)\,dx = \left[\frac{x^2}{2} - x\right]_{1}^{3} = \left(\frac{9}{2} - 3\right) - \left(\frac{1}{2} - 1\right) = \frac{3}{2} + \frac{1}{2} = 2$$

Step 4. Add.

$$8 + 2 = 10$$

Check­ing the answer#

The graph of $\lvert x - 1\rvert$ on $[-3, 3]$ is two tri­an­gles with their cor­ner at $x = 1$: the left one has base $4$ and height $4$ (area $8$), the right one base $2$ and height $2$ (area $2$). Total $10$.

Answer#

$\displaystyle \int_{-3}^{3}\lvert x - 1\rvert\,dx = 10$.

Ques­tion 5: Using odd and even func­tions#

The prob­lem#

Eval­u­ate: (a) $\displaystyle \int_{-\pi/2}^{\pi/2}\sin^7 x\,dx$ (b) $\displaystyle \int_{-1}^{1}(x^2 + x^3)\,dx$

Under­stand­ing the prob­lem#

Both inter­vals are sym­met­ric about $0$, of the form $[-a, a]$. That is the sig­nal to use prop­erty 5.

The idea#

Prop­erty 5: over $[-a, a]$, an odd func­tion ($f(-x) = -f(x)$) inte­grates to $0$, and an even func­tion ($f(-x) = f(x)$) gives $\displaystyle 2\int_0^a f$.

Step-by-step solu­tion#

Part (a)

Step 1. Test for odd or even: $\sin(-x) = -\sin x$, so $\sin^7(-x) = (-\sin x)^7 = -\sin^7 x$. The inte­grand is odd.

Step 2. By prop­erty 5, the inte­gral is $0$.

$$\displaystyle \int_{-\pi/2}^{\pi/2}\sin^7 x\,dx = 0$$

Part (b)

Step 1. Split into two inte­grals: $x^2$ is even, $x^3$ is odd.

$$\displaystyle \int_{-1}^{1}(x^2 + x^3)\,dx = \int_{-1}^{1}x^2\,dx + \int_{-1}^{1}x^3\,dx$$

Step 2. The odd part gives $0$; the even part gives twice the inte­gral from $0$ to $1$.

$$\displaystyle = 2\int_0^1 x^2\,dx + 0 = 2\left[\frac{x^3}{3}\right]_0^1 = \frac{2}{3}$$

Check­ing the answer#

(b) directly: $\displaystyle \left[\tfrac{x^3}{3} + \tfrac{x^4}{4}\right]_{-1}^{1} = \left(\tfrac{1}{3} + \tfrac{1}{4}\right) - \left(-\tfrac{1}{3} + \tfrac{1}{4}\right) = \tfrac{2}{3}$.

Answer#

(a) $0$; (b) $\displaystyle \frac{2}{3}$.

Ques­tion 6: The "replace $x$ by $\displaystyle \tfrac{\pi}{2} - x$ and add" method#

The prob­lem#

Eval­u­ate $\displaystyle \int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx$.

Under­stand­ing the prob­lem#

There is no easy anti­deriv­a­tive. But the inte­grand has a nice sym­me­try: replac­ing $x$ by $\displaystyle \tfrac{\pi}{2} - x$ swaps $\sin x$ and $\cos x$.

The idea#

Use prop­erty 4, $\displaystyle \int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$, with $\displaystyle a = \tfrac{\pi}{2}$, exactly as in Exam­ple 7. Adding the orig­i­nal and the new form gives an inte­grand equal to $1$.

Step-by-step solu­tion#

Step 1. Call the inte­gral $I$.

$$\displaystyle I = \int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx$$

Step 2. Replace $x$ by $\displaystyle \tfrac{\pi}{2} - x$. Since $\displaystyle \sin\left(\tfrac{\pi}{2} - x\right) = \cos x$ and $\displaystyle \cos\left(\tfrac{\pi}{2} - x\right) = \sin x$:

$$\displaystyle I = \int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx$$

Step 3. Add the two forms of $I$. The denom­i­na­tors are the same, so the numer­a­tors add.

$$\displaystyle 2I = \int_0^{\pi/2}\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}$$

Step 4. Divide by $2$.

$$\displaystyle I = \frac{\pi}{4}$$

Check­ing the answer#

The inte­grand rises from $0$ at $x = 0$ to $1$ at $\displaystyle x = \tfrac{\pi}{2}$, and equals $\displaystyle \tfrac{1}{2}$ at the mid­point. So the area should be about half the inter­val length, $\displaystyle \tfrac{1}{2}\cdot\tfrac{\pi}{2} = \tfrac{\pi}{4}$. It is.

Answer#

$\displaystyle I = \frac{\pi}{4}$.

Ques­tion 7: The same method with gen­eral lim­its#

The prob­lem#

Eval­u­ate $\displaystyle \int_2^8\frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}}\,dx$.

Under­stand­ing the prob­lem#

The lim­its are $2$ and $8$, so $a + b = 10$. Replac­ing $x$ by $10 - x$ swaps $\sqrt{x}$ and $\sqrt{10 - x}$.

The idea#

Use prop­erty 3, $\displaystyle \int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$, then add the two forms.

Step-by-step solu­tion#

Step 1. Call the inte­gral $I$.

$$\displaystyle I = \int_2^8\frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}}\,dx$$

Step 2. Replace $x$ by $2 + 8 - x = 10 - x$. Then $\sqrt{10 - x}$ becomes $\sqrt{x}$ and $\sqrt{x}$ becomes $\sqrt{10 - x}$.

$$\displaystyle I = \int_2^8\frac{\sqrt{x}}{\sqrt{10 - x} + \sqrt{x}}\,dx$$

Step 3. Add the two forms.

$$\displaystyle 2I = \int_2^8\frac{\sqrt{10 - x} + \sqrt{x}}{\sqrt{x} + \sqrt{10 - x}}\,dx = \int_2^8 1\,dx = 8 - 2 = 6$$

Step 4. Divide by $2$.

$$I = 3$$

Check­ing the answer#

At the mid­point $x = 5$ the inte­grand is $\displaystyle \tfrac{1}{2}$, and it is sym­met­ric about that value; so the answer is half the length of the inter­val: $\displaystyle \tfrac{1}{2} \times 6 = 3$.

Answer#

$I = 3$.

Ques­tion 8: The inte­gral of $\log(\sin x)$#

The prob­lem#

Eval­u­ate $\displaystyle \int_0^{\pi/2}\log(\sin x)\,dx$. The answer is $\displaystyle -\frac{\pi}{2}\log 2$; show the key step $\displaystyle 2I = \int_0^{\pi/2}\log\frac{\sin 2x}{2}\,dx$.

Under­stand­ing the prob­lem#

This inte­gral has no ele­men­tary anti­deriv­a­tive, so it must be found by prop­er­ties. You must show how the key step arises and how it leads to the answer. (The inte­grand becomes very neg­a­tive near $x = 0$, but the inte­gral still has a finite value.)

The idea#

Use prop­erty 4 to get a $\cos$ ver­sion, add the two ver­sions and use $\displaystyle \sin x\cos x = \frac{\sin 2x}{2}$. Then show that $\displaystyle \int_0^{\pi/2}\log(\sin 2x)\,dx$ is again equal to $I$, using a sub­sti­tu­tion and prop­erty 6.

Step-by-step solu­tion#

Step 1. Let $\displaystyle I = \int_0^{\pi/2}\log(\sin x)\,dx$. By prop­erty 4 (replace $x$ by $\displaystyle \tfrac{\pi}{2} - x$):

$$\displaystyle I = \int_0^{\pi/2}\log(\cos x)\,dx$$

Step 2. Add the two forms and use $\log A + \log B = \log(AB)$.

$$\displaystyle 2I = \int_0^{\pi/2}\log(\sin x\cos x)\,dx$$

Step 3. Use $\displaystyle \sin x\cos x = \frac{\sin 2x}{2}$. This is the key step asked for.

$$\displaystyle 2I = \int_0^{\pi/2}\log\frac{\sin 2x}{2}\,dx$$

Step 4. Split the log­a­rithm: $\displaystyle \log\frac{\sin 2x}{2} = \log(\sin 2x) - \log 2$.

$$\displaystyle 2I = \int_0^{\pi/2}\log(\sin 2x)\,dx - \frac{\pi}{2}\log 2$$

Step 5. Work out $\displaystyle J = \int_0^{\pi/2}\log(\sin 2x)\,dx$. Sub­sti­tute $t = 2x$, so $\displaystyle dx = \frac{dt}{2}$; the lim­its become $0$ and $\pi$.

$$\displaystyle J = \frac{1}{2}\int_0^{\pi}\log(\sin t)\,dt$$

Step 6. Since $\sin(\pi - t) = \sin t$, prop­erty 6 (with $2a = \pi$) gives $\displaystyle \int_0^{\pi}\log(\sin t)\,dt = 2\int_0^{\pi/2}\log(\sin t)\,dt = 2I$. So

$$\displaystyle J = \frac{1}{2}\cdot 2I = I$$

Step 7. Put this into Step 4 and solve for $I$.

$$\displaystyle 2I = I - \frac{\pi}{2}\log 2 \quad\Longrightarrow\quad I = -\frac{\pi}{2}\log 2$$

Check­ing the answer#

$\displaystyle -\tfrac{\pi}{2}\log 2 \approx -1.089$. The answer must be neg­a­tive, since $\sin x < 1$ makes $\log(\sin x) < 0$ on $\displaystyle \left(0, \tfrac{\pi}{2}\right)$. A numer­i­cal inte­gra­tion also gives about $-1.089$.

Answer#

$\displaystyle \int_0^{\pi/2}\log(\sin x)\,dx = -\frac{\pi}{2}\log 2$.

Ques­tion 9: A stan­dard form with a square root#

The prob­lem#

Eval­u­ate $\displaystyle \int_0^4\frac{dx}{\sqrt{x^2 + 9}}$.

Under­stand­ing the prob­lem#

The inte­grand has the form $\displaystyle \frac{1}{\sqrt{x^2 + a^2}}$ with $a = 3$. This is one of the stan­dard inte­grals from the chap­ter.

The idea#

Use $\displaystyle \int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$, then sub­sti­tute the lim­its.

Step-by-step solu­tion#

Step 1. Write the anti­deriv­a­tive with $a = 3$.

$$\displaystyle \int_0^4\frac{dx}{\sqrt{x^2 + 9}} = \Big[\log\left\lvert x + \sqrt{x^2 + 9}\right\rvert\Big]_0^4$$

Step 2. Upper limit: $4 + \sqrt{16 + 9} = 4 + 5 = 9$. Lower limit: $0 + \sqrt{9} = 3$.

$$= \log 9 - \log 3$$

Step 3. Com­bine with $\displaystyle \log A - \log B = \log\frac{A}{B}$.

$$\displaystyle = \log\frac{9}{3} = \log 3$$

Check­ing the answer#

$\log 3 \approx 1.099$. The inte­grand falls from $\displaystyle \tfrac{1}{3}$ to $\displaystyle \tfrac{1}{5}$ over an inter­val of length $4$, so the area is between $0.8$ and $1.33$. It is.

Answer#

$\log 3$.

Ques­tion 10: Par­tial frac­tions in a def­i­nite inte­gral#

The prob­lem#

Eval­u­ate $\displaystyle \int_1^3\frac{dx}{x(x + 1)}$.

Under­stand­ing the prob­lem#

The inte­grand is a ratio­nal func­tion whose denom­i­na­tor fac­torises into two lin­ear fac­tors. Split it into sim­pler frac­tions first.

The idea#

Write $\displaystyle \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$ (par­tial frac­tions), inte­grate each to a log­a­rithm, then use the lim­its.

Step-by-step solu­tion#

Step 1. Par­tial frac­tions: $\displaystyle \frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1}$ gives $1 = A(x + 1) + Bx$. Putting $x = 0$: $A = 1$. Putting $x = -1$: $B = -1$.

$$\displaystyle \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$$

Step 2. Inte­grate (on $[1, 3]$ every­thing is pos­i­tive).

$$\displaystyle \int_1^3\left(\frac{1}{x} - \frac{1}{x + 1}\right)dx = \big[\log x - \log(x + 1)\big]_1^3 = \left[\log\frac{x}{x + 1}\right]_1^3$$

Step 3. Sub­sti­tute the lim­its.

$$\displaystyle = \log\frac{3}{4} - \log\frac{1}{2}$$

Step 4. Com­bine: $\displaystyle \frac{3/4}{1/2} = \frac{3}{2}$.

$$\displaystyle = \log\frac{3}{2}$$

Check­ing the answer#

$\log 1.5 \approx 0.405$. The inte­grand falls from $0.5$ at $x = 1$ to $\displaystyle \tfrac{1}{12} \approx 0.083$ at $x = 3$; an area of about $0.4$ over length $2$ is rea­son­able.

Answer#

$\displaystyle \log\frac{3}{2}$.

Ques­tion 11: Inte­gra­tion by parts with an inverse func­tion#

The prob­lem#

Eval­u­ate $\displaystyle \int_0^1\tan^{-1}x\,dx$.

Under­stand­ing the prob­lem#

We know how to dif­fer­en­ti­ate $\tan^{-1}x$ but it is not in the table of inte­grals. The trick is to treat it as a prod­uct $\tan^{-1}x \times 1$.

The idea#

Inte­grate by parts with $u = \tan^{-1}x$ (its deriv­a­tive $\displaystyle \frac{1}{1 + x^2}$ is sim­pler) and $dv = dx$ (so $v = x$). The remain­ing inte­gral is then a sub­sti­tu­tion of the type in Ques­tion 2(a).

Step-by-step solu­tion#

Step 1. $u = \tan^{-1}x$, $\displaystyle du = \frac{dx}{1 + x^2}$; $dv = dx$, $v = x$.

$$\displaystyle \int_0^1\tan^{-1}x\,dx = \big[x\tan^{-1}x\big]_0^1 - \int_0^1\frac{x}{1 + x^2}\,dx$$

Step 2. First term: $\displaystyle 1\cdot\tan^{-1}1 - 0 = \frac{\pi}{4}$.

Step 3. Sec­ond term: this is Ques­tion 2(a), equal to $\displaystyle \frac{1}{2}\log 2$.

Step 4. Sub­tract.

$$\displaystyle \int_0^1\tan^{-1}x\,dx = \frac{\pi}{4} - \frac{1}{2}\log 2$$

Check­ing the answer#

$\displaystyle \tfrac{\pi}{4} - \tfrac{1}{2}\log 2 \approx 0.785 - 0.347 = 0.439$. The inte­grand rises from $0$ to $\displaystyle \tfrac{\pi}{4} \approx 0.785$ and bends down­ward, so the area should be a bit more than the tri­an­gle $\displaystyle \tfrac{1}{2}(0.785) \approx 0.39$. It is.

Answer#

$\displaystyle \frac{\pi}{4} - \frac{1}{2}\log 2$.

Ques­tion 12: The great­est inte­ger func­tion#

The prob­lem#

Eval­u­ate $\displaystyle \int_0^{2}[x]\,dx$, where $[x]$ is the great­est inte­ger less than or equal to $x$.

Under­stand­ing the prob­lem#

$[x]$ is a step func­tion: $[x] = 0$ for $0 \le x < 1$, and $[x] = 1$ for $1 \le x < 2$. Its value jumps at $x = 1$, so split there.

The idea#

Use prop­erty 2 to split $[0, 2]$ at $1$, and replace $[x]$ by its con­stant value on each piece. (A sin­gle point, such as $x = 2$ where $[x] = 2$, does not affect the inte­gral.)

Step-by-step solu­tion#

Step 1. Split at $x = 1$.

$$\displaystyle \int_0^2[x]\,dx = \int_0^1[x]\,dx + \int_1^2[x]\,dx$$

Step 2. On $[0, 1)$, $[x] = 0$.

$$\displaystyle \int_0^1 0\,dx = 0$$

Step 3. On $[1, 2)$, $[x] = 1$.

$$\displaystyle \int_1^2 1\,dx = 2 - 1 = 1$$

Step 4. Add.

$$0 + 1 = 1$$

Check­ing the answer#

The graph of $[x]$ on $[0, 2]$ is a flat line at height $0$ and then a flat line at height $1$ of width $1$: one unit square of area $1$.

Answer#

$\displaystyle \int_0^2[x]\,dx = 1$.