Definite integrals This is about the basics of evaluating definite integrals, and surprisingly how they can be used to define functions.
Contents Evaluation Suppose F ( x ) F(x) F ( x ) an antiderivative of f ( x ) f(x) f ( x )
∫ x 1 x 2 f ( x ) d x = F ( x 2 ) − F ( x 1 ) \int_{x_1}^{x_2} f(x)\d x = F(x_2) - F(x_1) ∫ x 1 x 2 f ( x ) d x = F ( x 2 ) − F ( x 1 ) It can be equivalently notated as
∫ x 1 x 2 f ( x ) d x = F ( x ) ∣ x 1 x 2 \int_{x_1}^{x_2} f(x)\d x = F(x)\Big\vert_{x_1}^{x_2} ∫ x 1 x 2 f ( x ) d x = F ( x ) x 1 x 2 And in cases to be absolutely clear, this can also be written as
∫ x 1 x 2 f ( x ) d x = F ( x ) ∣ x = x 1 x = x 2 \int_{x_1}^{x_2} f(x)\d x = F(x)\Big\vert_{x = x_1}^{x = x_2} ∫ x 1 x 2 f ( x ) d x = F ( x ) x = x 1 x = x 2 x 1 x_1 x 1 x 2 x_2 x 2 limits of integration .
Connecting back to indefinite integrals, we customarily set the + C +C + C 0 0 0
Example: Evaluate
∫ − e 2 − 1 1 x d x \int_{- \e^2}^{-1}\frac 1x \d x ∫ − e 2 − 1 x 1 d x ∫ − e 2 − 1 1 x d x = ln ∣ x ∣ ∣ − e 2 − 1 = ln ∣ − 1 ∣ − ln ∣ − e 2 ∣ = ln 1 − ln e 2 = 0 − 2 = − 2 ■ \begin{align*}\int_{- \e^2}^{-1}\frac 1x \d x &= \ln\lvert x\rvert \Big\vert_{- \e^2}^{-1} \\
&= \ln\lvert -1\rvert - \ln\lvert - \e^2\rvert \\
&= \ln 1 - \ln \e^2 \\
&= 0 - 2 = -2 \qed\end{align*} ∫ − e 2 − 1 x 1 d x = ln ∣ x ∣ − e 2 − 1 = ln ∣ − 1 ∣ − ln ∣ − e 2 ∣ = ln 1 − ln e 2 = 0 − 2 = − 2 ■ Where differentiation is about instantaneous change , definite integration is about the accumulation of change .
Definite integrals are defined as the signed area between a curve and the x x x . In this course, it is sufficient to use antiderivatives to find the area under a continuous function.
Properties These properties assume the function is defined for all x x x discontinuous (HL) at a finite number of points.
∫ x 1 x 2 f ( x ) d x = − ∫ x 2 x 1 f ( x ) d x \int_{x_1}^{x_2} f(x)\d x = -\int_{x_2}^{x_1} f(x)\d x ∫ x 1 x 2 f ( x ) d x = − ∫ x 2 x 1 f ( x ) d x ∫ x 1 x 2 f ( x ) d x + ∫ x 2 x 3 f ( x ) d x = ∫ x 1 x 3 f ( x ) d x \int_{x_1}^{x_2} f(x)\d x + \int_{x_2}^{x_3} f(x)\d x = \int_{x_1}^{x_3} f(x)\d x ∫ x 1 x 2 f ( x ) d x + ∫ x 2 x 3 f ( x ) d x = ∫ x 1 x 3 f ( x ) d x The properties from indefinite integrals also carry over.
∫ x 1 x 2 f ( x ) + g ( x ) d x = ∫ x 1 x 2 f ( x ) d x + ∫ x 1 x 2 g ( x ) d x \int_{x_1}^{x_2} f(x) + g(x)\d x = \int_{x_1}^{x_2} f(x) \d x + \int_{x_1}^{x_2} g(x) \d x ∫ x 1 x 2 f ( x ) + g ( x ) d x = ∫ x 1 x 2 f ( x ) d x + ∫ x 1 x 2 g ( x ) d x ∫ x 1 x 2 k f ( x ) d x = k ∫ x 1 x 2 f ( x ) d x \int_{x_1}^{x_2} k f(x)\d x = k\int_{x_1}^{x_2} f(x)\d x ∫ x 1 x 2 k f ( x ) d x = k ∫ x 1 x 2 f ( x ) d x With F ( x ) F(x) F ( x ) f ( x ) f(x) f ( x )
∫ x 1 x 2 f ( a x + b ) d x = 1 a ∫ a x 1 + b a x 2 + b f ( x ) d x = 1 a F ( a x 2 + b ) − 1 a F ( a x 1 + b ) \int_{x_1}^{x_2} f(ax+b)\d x = \frac 1a\int_{ax_1+b}^{ax_2+b} f(x)\d x = \frac 1a F(ax_2+b) - \frac 1a F(ax_1+b) ∫ x 1 x 2 f ( a x + b ) d x = a 1 ∫ a x 1 + b a x 2 + b f ( x ) d x = a 1 F ( a x 2 + b ) − a 1 F ( a x 1 + b ) The last property comes directly from substituting in x 1 x_1 x 1 x 2 x_2 x 2 u u u
Example: Find
∫ 0 1 ( 4 x + 2 ) 3 d x \int_0^1 (4x+2)^3 \d x ∫ 0 1 ( 4 x + 2 ) 3 d x Here we need to apply both integration of x n x^n x n f ( a x + b ) f(ax + b) f ( a x + b )
The antiderivative of x 3 x^3 x 3 1 4 x 4 + C \frac{1}{4}x^4 + C 4 1 x 4 + C
Then the f ( 4 x + 2 ) f(4x + 2) f ( 4 x + 2 ) 4 4 4
∫ 0 1 ( 4 x + 2 ) 3 d x = 1 16 ( 4 x + 2 ) 4 ∣ 0 1 = 1 2 4 ( 6 4 − 2 4 ) = 3 4 − 1 = 80 ■ \begin{align*}
\int_0^1 (4x+2)^3 \d x &= \frac {1}{16}(4x+2)^4 \Big\vert_0^1 \\
&= \frac{1}{2^4}(6^4 - 2^4) \\
&= 3^4 - 1 \\
&= 80 \qed
\end{align*} ∫ 0 1 ( 4 x + 2 ) 3 d x = 16 1 ( 4 x + 2 ) 4 0 1 = 2 4 1 ( 6 4 − 2 4 ) = 3 4 − 1 = 80 ■ Alternatively we could have factored out a 2 3 2^3 2 3
∫ 0 1 ( 4 x + 2 ) 3 d x = 8 ∫ 0 1 ( 2 x + 1 ) 3 d x = 8 ⋅ 1 2 ⋅ 4 [ ( 2 x + 1 ) 4 ] 0 1 = 81 − 1 = 80 ■ \begin{align*}
\int_0^1 (4x+2)^3 \d x &= 8\int_0^1 (2x+1)^3 \d x \\
&= 8\cdot\frac{1}{2\cdot 4}\left[(2x+1)^4\right]_0^1 \\
&= 81 - 1 \\
&= 80 \qed
\end{align*} ∫ 0 1 ( 4 x + 2 ) 3 d x = 8 ∫ 0 1 ( 2 x + 1 ) 3 d x = 8 ⋅ 2 ⋅ 4 1 [ ( 2 x + 1 ) 4 ] 0 1 = 81 − 1 = 80 ■ which in retrospect is easier, but not by a lot.
Variable of integration A key difference between indefinite and definite integration is that indefinite integrals return families of functions, but definite integration return a value. Just like how in f ( x ) f(x) f ( x ) x x x x x x
As such all of the following are equivalent
∫ − 3 5 f ( x ) d x \int_{-3}^{5} f(x)\d x ∫ − 3 5 f ( x ) d x ∫ − 3 5 f ( w ) d w \int_{-3}^{5} f(w)\d w ∫ − 3 5 f ( w ) d w ∫ − 3 5 f ( τ ) d τ \int_{-3}^{5} f(\tau)\d \tau ∫ − 3 5 f ( τ ) d τ Here w w w x x x τ \tau τ dummy variables or variables of integration .
Note to HL candidates: for u u u w = x w = x w = x τ = x \tau = x τ = x
Using definite integral to define a function There is a way to specify the antiderivative function F ( x ) F(x) F ( x )
F ( x ) = ∫ x 1 x f ( τ ) d τ F(x) = \int_{x_1}^{x} f(\tau)\d \tau F ( x ) = ∫ x 1 x f ( τ ) d τ On a calculator, the solver should be used; graphing takes too long for definite integrals.
Example: Simplify
f ( x ) = ∫ 0 x cos 2 θ d θ f(x) = \int_{0}^{x} \cos^2\theta\d \theta f ( x ) = ∫ 0 x cos 2 θ d θ Use double angle identity for cos 2 θ \cos 2\theta cos 2 θ cos 2 θ \cos^2\theta cos 2 θ
cos 2 θ = 2 cos 2 θ − 1 1 2 + 1 2 cos 2 θ = cos 2 θ \begin{align*}\cos 2\theta &= 2\cos^2\theta - 1 \\
\frac12 + \frac12\cos 2\theta &= \cos^2\theta \end{align*} cos 2 θ 2 1 + 2 1 cos 2 θ = 2 cos 2 θ − 1 = cos 2 θ We find the antiderivative, using the + C = 0 +C = 0 + C = 0
f ( x ) = ∫ 0 x 1 2 + 1 2 cos 2 θ d θ = [ 1 2 θ + 1 4 sin 2 θ ] θ = 0 θ = x = 1 2 x + 1 4 sin 2 x − ( 1 2 ( 0 ) + 1 4 sin 2 ( 0 ) ) = 1 2 x + 1 4 sin 2 x ■ \begin{align*}f(x) &= \int_0^x \frac12 + \frac12\cos 2\theta \d\theta \\
&= \left[\frac12\theta + \frac14\sin2\theta \right]_{\theta=0}^{\theta=x} \\
&= \frac12x + \frac14\sin2x - \left(\frac12(0) + \frac14\sin2(0)\right) \\
&= \frac12x + \frac14\sin2x \qed\end{align*} f ( x ) = ∫ 0 x 2 1 + 2 1 cos 2 θ d θ = [ 2 1 θ + 4 1 sin 2 θ ] θ = 0 θ = x = 2 1 x + 4 1 sin 2 x − ( 2 1 ( 0 ) + 4 1 sin 2 ( 0 ) ) = 2 1 x + 4 1 sin 2 x ■ In this particular question the second part entirely cancels. This should not assumed to be the case.
See an example on TI-84 Plus.
Derivation: z-scores of normal distribution Derivation: Let
f ( x ) = 2 π ∫ 0 x e − t 2 d t f(x) = \frac{2}{\sqrt\pi} \int_0^x \e^{-t^2} \d t f ( x ) = π 2 ∫ 0 x e − t 2 d t and given the normal distribution probability density function
g ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 g(x) = \frac{1}{\sigma\sqrt{2\pi}} \e^{-\frac{(x-\mu)^2}{2\sigma^2}} g ( x ) = σ 2 π 1 e − 2 σ 2 ( x − μ ) 2 Show that
∫ μ μ + z σ g ( x ) d x = 1 2 f ( z 2 ) for all z ∈ R \int_{\mu}^{\mu+z\sigma} g(x) \d x = \frac12f\left(\frac{z}{\sqrt2}\right) \text{ for all } z\in\mathbb R ∫ μ μ + z σ g ( x ) d x = 2 1 f ( 2 z ) for all z ∈ R I = 1 σ 2 π ∫ μ μ + z σ e − ( x − μ ) 2 2 σ 2 d x = 1 σ 2 π ∫ μ μ + z σ e − ( x − μ σ 2 ) 2 d x = σ 2 σ 2 π ∫ μ − μ σ 2 μ + z σ − μ σ 2 e − t 2 d t = 1 π ∫ 0 z 2 e − t 2 d t = 1 2 ( f ( z 2 ) − f ( 0 ) ) = 1 2 f ( z 2 ) ■ \begin{align*}
I &= \frac{1}{\sigma\sqrt{2\pi}} \int_{\mu}^{\mu+z\sigma} \e^{-\frac{(x-\mu)^2}{2\sigma^2}} \d x \\
&= \frac{1}{\sigma\sqrt{2\pi}} \int_{\mu}^{\mu+z\sigma} \e^{-\left(\frac{x-\mu}{\sigma\sqrt2}\right)^2} \d x \\
&= \frac{\sigma\sqrt2}{\sigma\sqrt{2\pi}} \int_{\frac{\mu - \mu}{\sigma\sqrt2}}^{\frac{\mu + z\sigma - \mu}{\sigma\sqrt2}} \e^{-t^2} \d t \\
&= \frac{1}{\sqrt\pi} \int_{0}^{\frac{z}{\sqrt2}} \e^{-t^2} \d t \\
&= \frac12 \left(f\left(\frac{z}{\sqrt2}\right) - f(0) \right) \\
&= \frac12f\left(\frac{z}{\sqrt2}\right) \qed
\end{align*} I = σ 2 π 1 ∫ μ μ + z σ e − 2 σ 2 ( x − μ ) 2 d x = σ 2 π 1 ∫ μ μ + z σ e − ( σ 2 x − μ ) 2 d x = σ 2 π σ 2 ∫ σ 2 μ − μ σ 2 μ + z σ − μ e − t 2 d t = π 1 ∫ 0 2 z e − t 2 d t = 2 1 ( f ( 2 z ) − f ( 0 ) ) = 2 1 f ( 2 z ) ■ f f f
We made use of
∫ x 1 x 2 f ( a x + b ) d x = 1 a ∫ a x 1 + b a x 2 + b f ( x ) d x \int_{x_1}^{x_2} f(ax+b)\d x = \frac 1a\int_{ax_1+b}^{ax_2+b} f(x)\d x ∫ x 1 x 2 f ( a x + b ) d x = a 1 ∫ a x 1 + b a x 2 + b f ( x ) d x and
∫ 0 0 f ( x ) d x = 0 \int_0^0 f(x)\d x = 0 ∫ 0 0 f ( x ) d x = 0 With x = μ + z σ x = \mu + z\sigma x = μ + z σ
z = x − μ σ z = \frac{x - \mu}{\sigma} z = σ x − μ In other words, probability from mean to z z z z z z
Notes on trig Calculus of trigonometric functions are only available in radians. Ensure that your calculator is in radians mode.