5.5 Substitution (Page 3/6)

Calculus volume 1 Page 3 / 6

\int f (g (x)) g^{'} (x) d x = F (g (x)) + C .

Then

\begin{matrix} \int_{a}^{b} f [g (x)] g^{'} (x) d x & = {F (g (x)) |}_{x = a}^{x = b} \\ = F (g (b)) - F (g (a)) \\ = {F (u) |}_{u = g (a)}^{u = g (b)} \\ = \int_{g (a)}^{g (b)} f (u) d u, \end{matrix}

and we have the desired result.

Using substitution to evaluate a definite integral

Use substitution to evaluate $\int_{0}^{1} x^{2} {(1 + 2 x^{3})}^{5} d x .$

Let $u = 1 + 2 x^{3},$ so $d u = 6 x^{2} d x .$ Since the original function includes one factor of x ² and $d u = 6 x^{2} d x,$ multiply both sides of the du equation by $1 / 6 .$ Then,

\begin{matrix} d u & = & 6 x^{2} d x \\ \frac{1}{6} d u & = & x^{2} d x . \end{matrix}

To adjust the limits of integration, note that when $x = 0, u = 1 + 2 (0) = 1,$ and when $x = 1, u = 1 + 2 (1) = 3 .$ Then

\int_{0}^{1} x^{2} {(1 + 2 x^{3})}^{5} d x = \frac{1}{6} \int_{1}^{3} u^{5} d u .

Evaluating this expression, we get

\begin{matrix} \frac{1}{6} \int_{1}^{3} u^{5} d u & = (\frac{1}{6}) (\frac{u^{6}}{6}) |_{1}^{3} \\ = \frac{1}{36} [{(3)}^{6} - {(1)}^{6}] \\ = \frac{182}{9} . \end{matrix}

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Use substitution to evaluate the definite integral $\int_{−1}^{0} y {(2 y^{2} - 3)}^{5} d y .$

$\frac{91}{3}$

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Using substitution with an exponential function

Use substitution to evaluate $\int_{0}^{1} x e^{4 x^{2} + 3} d x .$

Let $u = 4 x^{3} + 3 .$ Then, $d u = 8 x d x .$ To adjust the limits of integration, we note that when $x = 0, u = 3,$ and when $x = 1, u = 7 .$ So our substitution gives

\begin{matrix} \int_{0}^{1} x e^{4 x^{2} + 3} d x & = \frac{1}{8} \int_{3}^{7} e^{u} d u \\ = \frac{1}{8} e^{u} |_{3}^{7} \\ = \frac{e^{7} - e^{3}}{8} \\ \approx 134.568. \end{matrix}

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Use substitution to evaluate $\int_{0}^{1} x^{2} cos (\frac{π}{2} x^{3}) d x .$

$\frac{2}{3 π} \approx 0.2122$

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Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in [link] .

Using substitution to evaluate a trigonometric integral

Use substitution to evaluate $\int_{0}^{π / 2} {cos}^{2} θ d θ .$

Let us first use a trigonometric identity to rewrite the integral. The trig identity ${cos}^{2} θ = \frac{1 + cos 2 θ}{2}$ allows us to rewrite the integral as

\int_{0}^{π / 2} {cos}^{2} θ d θ = \int_{0}^{π / 2} \frac{1 + cos 2 θ}{2} d θ .

Then,

\begin{matrix} \int_{0}^{π / 2} (\frac{1 + cos 2 θ}{2}) d θ & = \int_{0}^{π / 2} (\frac{1}{2} + \frac{1}{2} cos 2 θ) d θ \\ = \frac{1}{2} \int_{0}^{π / 2} d θ + \int_{0}^{π / 2} cos 2 θ d θ . \end{matrix}

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let $u = 2 θ .$ Then, $d u = 2 d θ,$ or $\frac{1}{2} d u = d θ .$ Also, when $θ = 0, u = 0,$ and when $θ = π / 2, u = π .$ Expressing the second integral in terms of u , we have

\begin{matrix} \frac{1}{2} \int_{0}^{π / 2} d θ + \frac{1}{2} \int_{0}^{π / 2} cos 2 θ d θ & = \frac{1}{2} \int_{0}^{π / 2} d θ + \frac{1}{2} (\frac{1}{2}) \int_{0}^{π} cos u d u \\ = \frac{θ}{2} |_{θ = 0}^{θ = π / 2} + \frac{1}{4} sin u |_{u = 0}^{u = θ} \\ = (\frac{π}{4} - 0) + (0 - 0) = \frac{π}{4} . \end{matrix}

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Key concepts

Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions in the integrand.
When using substitution for a definite integral, we also have to change the limits of integration.

Key equations

Substitution with Indefinite Integrals
$\int f [g (x)] g^{'} (x) d x = \int f (u) d u = F (u) + C = F (g (x)) + C$
Substitution with Definite Integrals
$\int_{a}^{b} f (g (x)) g' (x) d x = \int_{g (a)}^{g (b)} f (u) d u$

Why is u -substitution referred to as change of variable ?

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2. If $f = g \circ h,$ when reversing the chain rule, $\frac{d}{d x} (g \circ h) (x) = g^{'} (h (x)) h^{'} (x),$ should you take $u = g (x)$ or $u = h (x) ?$

$u = h (x)$

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In the following exercises, verify each identity using differentiation. Then, using the indicated u -substitution, identify f such that the integral takes the form $\int f (u) d u .$

$\int x \sqrt{x + 1} d x = \frac{2}{15} {(x + 1)}^{3 / 2} (3 x - 2) + C; u = x + 1$

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$\int \frac{x^{2}}{\sqrt{x - 1}} d x (x > 1) = \frac{2}{15} \sqrt{x - 1} (3 x^{2} + 4 x + 8) + C; u = x - 1$

$f (u) = \frac{{(u + 1)}^{2}}{\sqrt{u}}$

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Questions & Answers

find the equation of the tangent to the curve y=2x³-x²+3x+1 at the points x=1 and x=3

Esther Reply

derivative of logarithms function

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ex 2.1 question no 11

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ex 2.1 question no. 11

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Find the derivative of g(x)=−3.

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abdulkadir

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)

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fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5

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how to use fundamental theorem to solve exponential

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Chicken nuggets

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🐔🦃 nuggets

(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.

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log tan (x/4+x/2)

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y=(x^2 + 3x).(eipix)

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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