The ∫{\displaystyle \int } is the symbol for integration. It is actually an elongated S. The function f(x){\displaystyle f(x)} is called the integrand when it is inside the integral. The differential dx{\displaystyle \mathrm {d} x} intuitively is saying what variable you are integrating with respect to. Because (Riemann) integration is just a sum of infinitesimally thin rectangles with a height of f(x),{\displaystyle f(x),} we see that dx{\displaystyle \mathrm {d} x} refers to the width of those rectangles. The letters a{\displaystyle a} and b{\displaystyle b} are the boundaries. An integral does not need to have boundaries. When this is the case, we say that we are dealing with an indefinite integral. If it does, then we are dealing with a definite integral. Throughout this article, we will go over the process of finding antiderivatives of a function. An antiderivative is a function whose derivative is the original function we started with.
∑i=1nf(xi)Δxi{\displaystyle \sum {i=1}^{n}f(x{i})\Delta x_{i}} If this is the first time you have seen a summation symbol, it may look scary. . . but it’s not complicated at all. All this is saying is that we are summing up the area of n{\displaystyle n} rectangles. (The variable i{\displaystyle i} is known as a dummy index. ) However, as you can guess, the area of all the rectangles is bound to be slightly different from the true area. We solve this by sending the number of rectangles to infinity. As we increase the number of rectangles, the area of all the rectangles better approximates the area under the curve. That’s what the diagram above shows (see the tips for what the graph in the middle shows). The limit as n→∞{\displaystyle n\to \infty } is what we define as the integral of the function f(x){\displaystyle f(x)} from a{\displaystyle a} to b. {\displaystyle b. } ∫abf(x)dx=limn→∞∑i=1nf(xi)Δxi{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x=\lim {n\to \infty }\sum {i=1}^{n}f(x{i})\Delta x{i}} Of course, this limit has to exist in order for the integral to have any meaning. If such a limit does not exist on the interval, then we say that f(x){\displaystyle f(x)} does not have an integral over the interval [a,b]. {\displaystyle [a,b]. } In this article (and in almost every physical application), we only deal with functions where these integrals exist.
∑i=1nf(xi)Δxi{\displaystyle \sum {i=1}^{n}f(x{i})\Delta x_{i}} If this is the first time you have seen a summation symbol, it may look scary. . . but it’s not complicated at all. All this is saying is that we are summing up the area of n{\displaystyle n} rectangles. (The variable i{\displaystyle i} is known as a dummy index. ) However, as you can guess, the area of all the rectangles is bound to be slightly different from the true area. We solve this by sending the number of rectangles to infinity. As we increase the number of rectangles, the area of all the rectangles better approximates the area under the curve. That’s what the diagram above shows (see the tips for what the graph in the middle shows). The limit as n→∞{\displaystyle n\to \infty } is what we define as the integral of the function f(x){\displaystyle f(x)} from a{\displaystyle a} to b. {\displaystyle b. } ∫abf(x)dx=limn→∞∑i=1nf(xi)Δxi{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x=\lim {n\to \infty }\sum {i=1}^{n}f(x{i})\Delta x{i}} Of course, this limit has to exist in order for the integral to have any meaning. If such a limit does not exist on the interval, then we say that f(x){\displaystyle f(x)} does not have an integral over the interval [a,b]. {\displaystyle [a,b]. } In this article (and in almost every physical application), we only deal with functions where these integrals exist.
∫xndx=xn+1n+1+C{\displaystyle \int x^{n}\mathrm {d} x={\frac {x^{n+1}}{n+1}}+C} To verify that this power rule holds, differentiate the antiderivative to recover the original function. The power rule holds for all functions of this form with degree n{\displaystyle n} except when n=−1. {\displaystyle n=-1. } We will see why later.
∫(axn+bxm)dx=a∫xndx+b∫xmdx{\displaystyle \int (ax^{n}+bx^{m})\mathrm {d} x=a\int x^{n}\mathrm {d} x+b\int x^{m}\mathrm {d} x} This should be familiar because the derivative is also a linear operator; the derivative of a sum is the sum of the derivatives. Linearity does not apply just for integrals of polynomials. It applies to any integral where the integrand is a sum of two or more terms.
∫(x4+2x3−5x2−1)dx=15x5+24x4−53x3−x+C{\displaystyle \int (x^{4}+2x^{3}-5x^{2}-1)\mathrm {d} x={\frac {1}{5}}x^{5}+{\frac {2}{4}}x^{4}-{\frac {5}{3}}x^{3}-x+C}
∫f(x)dx=∫2x2+3x−1x1/3dx=∫(2x2x1/3+3xx1/3−1x1/3)dx=∫(2x5/3+3x2/3−x−1/3)dx=2⋅38x8/3+3⋅35x5/3−32x2/3+C=34x8/3+95x5/3−32x2/3+C{\displaystyle {\begin{aligned}\int f(x)\mathrm {d} x&=\int {\frac {2x^{2}+3x-1}{x^{1/3}}}\mathrm {d} x\&=\int \left({\frac {2x^{2}}{x^{1/3}}}+{\frac {3x}{x^{1/3}}}-{\frac {1}{x^{1/3}}}\right)\mathrm {d} x\&=\int (2x^{5/3}+3x^{2/3}-x^{-1/3})\mathrm {d} x\&=2\cdot {\frac {3}{8}}x^{8/3}+3\cdot {\frac {3}{5}}x^{5/3}-{\frac {3}{2}}x^{2/3}+C\&={\frac {3}{4}}x^{8/3}+{\frac {9}{5}}x^{5/3}-{\frac {3}{2}}x^{2/3}+C\end{aligned}}} The common theme is that you must perform whatever manipulations in order to get the integral into a polynomial. From there, integration is easy. Judging whether the integral is easy enough to brute-force, or requires some algebraic manipulation first, is where the skill lies.
∫23x2dx{\displaystyle \int _{2}^{3}x^{2}\mathrm {d} x}
Let F(x){\displaystyle F(x)} be an antiderivative of f(x). {\displaystyle f(x). } Then ∫abf(x)dx=F(b)−F(a). {\displaystyle \int _{a}^{b}f(x)\mathrm {d} x=F(b)-F(a). } This theorem is incredibly useful because it simplifies the integral and means that the definite integral is completely determined by just the values on its boundaries. There is no need to sum up rectangles anymore to compute integrals. All we need to do now is to find antiderivatives, and evaluate at the bounds!
∫23x2dx=13x3|23=13(3)3−13(2)3=193{\displaystyle {\begin{aligned}\int {2}^{3}x^{2}\mathrm {d} x&={\frac {1}{3}}x^{3}{\Bigg |}{2}^{3}\&={\frac {1}{3}}(3)^{3}-{\frac {1}{3}}(2)^{3}\&={\frac {19}{3}}\end{aligned}}} Again, the fundamental theorem of calculus does not just apply to functions like f(x)=x2. {\displaystyle f(x)=x^{2}. } The fundamental theorem can be used to integrate any function, as long as you can find an antiderivative.
∫32x2dx=13x3|32=13(2)3−13(3)3=−193{\displaystyle {\begin{aligned}\int {3}^{2}x^{2}\mathrm {d} x&={\frac {1}{3}}x^{3}{\Bigg |}{3}^{2}\&={\frac {1}{3}}(2)^{3}-{\frac {1}{3}}(3)^{3}\&=-{\frac {19}{3}}\end{aligned}}} We just obtained the negative of the answer we got before. This illustrates an important property of definite integrals. Swapping the boundaries negates the integral. ∫baf(x)dx=−∫abf(x)dx{\displaystyle \int _{b}^{a}f(x)\mathrm {d} x=-\int _{a}^{b}f(x)\mathrm {d} x}
∫exdx=ex+C{\displaystyle \int e^{x}\mathrm {d} x=e^{x}+C} ∫axdx=axlna+C{\displaystyle \int a^{x}\mathrm {d} x={\frac {a^{x}}{\ln a}}+C}
∫sinxdx=−cosx+C{\displaystyle \int \sin x\mathrm {d} x=-\cos x+C} ∫cosxdx=sinx+C{\displaystyle \int \cos x\mathrm {d} x=\sin x+C} ∫sec2xdx=tanx+C{\displaystyle \int \sec ^{2}x\mathrm {d} x=\tan x+C} ∫csc2xdx=−cotx+C{\displaystyle \int \csc ^{2}x\mathrm {d} x=-\cot x+C} ∫secxtanxdx=secx+C{\displaystyle \int \sec x\tan x\mathrm {d} x=\sec x+C} ∫cscxcotxdx=−cscx+C{\displaystyle \int \csc x\cot x\mathrm {d} x=-\csc x+C}
∫11−x2dx=sin−1x+C{\displaystyle \int {\frac {1}{\sqrt {1-x^{2}}}}\mathrm {d} x=\sin ^{-1}x+C} ∫−11−x2dx=cos−1x+C{\displaystyle \int {\frac {-1}{\sqrt {1-x^{2}}}}\mathrm {d} x=\cos ^{-1}x+C} ∫11+x2dx=tan−1x+C{\displaystyle \int {\frac {1}{1+x^{2}}}\mathrm {d} x=\tan ^{-1}x+C}
∫1xdx=ln|x|+C{\displaystyle \int {\frac {1}{x}}\mathrm {d} x=\ln |x|+C} (Sometimes, authors like to put the dx{\displaystyle \mathrm {d} x} in the numerator of the fraction, so it reads like ∫dxx. {\displaystyle \int {\frac {\mathrm {d} x}{x}}. } Be aware of this notation. ) The reason for the absolute value in the logarithm function is subtle, and requires a more thorough understanding of real analysis in order to fully answer. For now, we will just live with the fact that the domains become the same when the absolute value bars are added.
∫π/4π/3f(x)dx=∫π/4π/3(2cosx+tan2x−6)dx=∫π/4π/3(2cosx+sec2x−7)dx=(2sinx+tanx−7x)|π/4π/3=(2sinπ3+tanπ3−7π3)−(2sinπ4+tanπ4−7π4)=23−2−1−7π12{\displaystyle {\begin{aligned}\int _{\pi /4}^{\pi /3}f(x)\mathrm {d} x&=\int _{\pi /4}^{\pi /3}(2\cos x+\tan ^{2}x-6)\mathrm {d} x\&=\int {\pi /4}^{\pi /3}(2\cos x+\sec ^{2}x-7)\mathrm {d} x\&=(2\sin x+\tan x-7x){\Big |}{\pi /4}^{\pi /3}\&=\left(2\sin {\frac {\pi }{3}}+\tan {\frac {\pi }{3}}-{\frac {7\pi }{3}}\right)-\left(2\sin {\frac {\pi }{4}}+\tan {\frac {\pi }{4}}-{\frac {7\pi }{4}}\right)\&=2{\sqrt {3}}-{\sqrt {2}}-1-{\frac {7\pi }{12}}\end{aligned}}} If you need a decimal approximation, you can use a calculator. Here, 23−2−1−7π12≈−0. 7827. {\displaystyle 2{\sqrt {3}}-{\sqrt {2}}-1-{\frac {7\pi }{12}}\approx -0. 7827. }
∫−11(cosx+x4)dx{\displaystyle \int _{-1}^{1}(\cos x+x^{4})\mathrm {d} x} Our integrand is even. We can immediately integrate by using the fundamental theorem of calculus, but if we look more carefully, we see that the bounds are symmetric about x=0. {\displaystyle x=0. } That means that the integral from -1 to 0 is going to give us the same value as the integral from 0 to 1. So what we can do is we can change the bounds to 0 and 1 and factor out a 2. ∫−11(cosx+x4)dx=2∫01(cosx+x4)dx{\displaystyle \int _{-1}^{1}(\cos x+x^{4})\mathrm {d} x=2\int _{0}^{1}(\cos x+x^{4})\mathrm {d} x} It might not seem like much to do this, but we will immediately see that our work is simplified. After finding the antiderivative, notice that we only need to evaluate it at x=1. {\displaystyle x=1. } The antiderivative at x=0{\displaystyle x=0} will not contribute to the integral. 2∫01(cosx+x4)dx=2sin1+25{\displaystyle 2\int _{0}^{1}(\cos x+x^{4})\mathrm {d} x=2\sin 1+{\frac {2}{5}}} In general, whenever you see an even function with symmetric boundaries, you should perform this simplification in order to make less arithmetic mistakes. ∫−aaf(x)dx=2∫0af(x)dx, f(x) even{\displaystyle \int _{-a}^{a}f(x)\mathrm {d} x=2\int _{0}^{a}f(x)\mathrm {d} x,\ \ f(x)\ \mathrm {even} }
∫−π/2π/2(2x3+2sinx)dx{\displaystyle \int _{-\pi /2}^{\pi /2}(2x^{3}+2\sin x)\mathrm {d} x} We could evaluate this integral directly. . . or we can recognize that our integrand is odd. Furthermore, the boundaries are symmetric about the origin. Therefore, our integral is 0. Why is this the case? It is because the antiderivative is even. Even functions have the property that f(−x)=f(x),{\displaystyle f(-x)=f(x),} so when we evaluate at the bounds −a{\displaystyle -a} and a,{\displaystyle a,} then F(−a)=F(a){\displaystyle F(-a)=F(a)} immediately implies that F(a)−F(−a)=0. {\displaystyle F(a)-F(-a)=0. } ∫−π/2π/2(2x3+2sinx)dx=0{\displaystyle \int _{-\pi /2}^{\pi /2}(2x^{3}+2\sin x)\mathrm {d} x=0} The properties of these functions are very potent in simplifying the integrals, but the boundaries must be symmetric. Otherwise, we will need to evaluate the old way. ∫−aaf(x)dx=0, f(x) odd{\displaystyle \int _{-a}^{a}f(x)\mathrm {d} x=0,\ \ f(x)\ \mathrm {odd} }
∫eaxdx{\displaystyle \int e^{ax}\mathrm {d} x}
∫eaxdx=1a∫eudu{\displaystyle \int e^{ax}\mathrm {d} x={\frac {1}{a}}\int e^{u}\mathrm {d} u}
∫eaxdx=1aeax+C{\displaystyle \int e^{ax}\mathrm {d} x={\frac {1}{a}}e^{ax}+C}
∫01x2x+3dx{\displaystyle \int _{0}^{1}x{\sqrt {2x+3}}\mathrm {d} x}
∫01x2x+3dx=12∫35xudu{\displaystyle \int _{0}^{1}x{\sqrt {2x+3}}\mathrm {d} x={\frac {1}{2}}\int _{3}^{5}x{\sqrt {u}}\mathrm {d} u}
12∫35xudu=14∫35(u−3)udu{\displaystyle {\frac {1}{2}}\int _{3}^{5}x{\sqrt {u}}\mathrm {d} u={\frac {1}{4}}\int _{3}^{5}(u-3){\sqrt {u}}\mathrm {d} u}
14∫35(u−3)udu=14∫35(u3/2−3u1/2)du=14(25u5/2−3⋅23u3/2)|35=14(25(5)5/2−2⋅(5)3/2−25(3)5/2+2⋅(3)3/2)=14(−65⋅33/2+2⋅33/2)=144533/2=335{\displaystyle {\begin{aligned}{\frac {1}{4}}\int _{3}^{5}(u-3){\sqrt {u}}\mathrm {d} u&={\frac {1}{4}}\int {3}^{5}(u^{3/2}-3u^{1/2})\mathrm {d} u\&={\frac {1}{4}}\left({\frac {2}{5}}u^{5/2}-3\cdot {\frac {2}{3}}u^{3/2}\right){\Bigg |}{3}^{5}\&={\frac {1}{4}}\left({\frac {2}{5}}(5)^{5/2}-2\cdot (5)^{3/2}-{\frac {2}{5}}(3)^{5/2}+2\cdot (3)^{3/2}\right)\&={\frac {1}{4}}\left(-{\frac {6}{5}}\cdot 3^{3/2}+2\cdot 3^{3/2}\right)\&={\frac {1}{4}}{\frac {4}{5}}3^{3/2}\&={\frac {3{\sqrt {3}}}{5}}\end{aligned}}}
∫udv=uv−∫vdu{\displaystyle \int u\mathrm {d} v=uv-\int v\mathrm {d} u}
∫lnxdx{\displaystyle \int \ln x\mathrm {d} x}
∫lnxdx=xlnx−∫dx{\displaystyle \int \ln x\mathrm {d} x=x\ln x-\int \mathrm {d} x} We converted the integral of a logarithm into the integral of 1, which is trivial to evaluate.
∫lnxdx=xlnx−x+C{\displaystyle \int \ln x\mathrm {d} x=x\ln x-x+C}
