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mcastillo356

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- TL;DR Summary
- Need some help understanding the so-called Theorem.

Hi, PF

I am going to reproduce the introduction of the textbook; then the Theorem:

The method of substitution cannot be forced to work. There is no substitution that will do much good with the integral ##\int{x(2+x^7)^{1/5}}\,dx##, for instance. However, the integral ##\int{x^6(2+x^7)^{1/5}}\,dx## will yield to the substitution ##u=2+x^7##. The substitution ##u=g(x)## is more likely to work if ##g'(x)## is a factor of the integrand.

The following theorem simplifies the use of the method of susbstitution in definite integrals.

THEOREM 6 Substitution in a definite integral

Suppose that ##g## is a differentiable function on ##[a,b]## that satisfies ##g(a)=A## and ##g(b)=B##. Also suppose that ##f## is continous on the range of ##g##. Then

$$\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}$$

PROOF Let ##F## be an antiderivative of ##f: F'(u)=f(u)##. Then

##\displaystyle\frac{d}{dx}\,F(g(x))=F'(g(x))g'(x)##.

Thus,

$$\displaystyle\int_a^b{f(g(x))g'(x)}=F(g(x))\Bigg |_a^b=F(g(b))-F(g(a))

=F(B)-F(A)=F(u)\Bigg |_A^B=\displaystyle\int_A^B{\,f(u)du}$$

EXAMPLE 5 Evaluate the integral ##I=\displaystyle\int_0^8{\displaystyle\frac{\cos{\sqrt{x+1}}}{\sqrt{x+1}}\,dx}##.

Solution Let ##u=\sqrt{x+1}##. Then ##du=\displaystyle\frac{dx}{2\sqrt{x+1}}##. If ##x=0##, then ##u=1##; if ##x=8##, then ##u=3##. Thus

$$I=2\displaystyle\int_1^3{\cos{u}\,du}=2\sin{u\Bigg |_1^3}=2\sin{3}-2\sin{1}$$

Paradoxically, I understand the example 5, but can't deal with the theory, ie, the theorem 6. Try to focus the doubt about the theorem will be my attempt: why in the equality ##\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}##, at the left, the integrand is a composite function?.

Greetings

PD: Post without preview

I am going to reproduce the introduction of the textbook; then the Theorem:

The method of substitution cannot be forced to work. There is no substitution that will do much good with the integral ##\int{x(2+x^7)^{1/5}}\,dx##, for instance. However, the integral ##\int{x^6(2+x^7)^{1/5}}\,dx## will yield to the substitution ##u=2+x^7##. The substitution ##u=g(x)## is more likely to work if ##g'(x)## is a factor of the integrand.

The following theorem simplifies the use of the method of susbstitution in definite integrals.

THEOREM 6 Substitution in a definite integral

Suppose that ##g## is a differentiable function on ##[a,b]## that satisfies ##g(a)=A## and ##g(b)=B##. Also suppose that ##f## is continous on the range of ##g##. Then

$$\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}$$

PROOF Let ##F## be an antiderivative of ##f: F'(u)=f(u)##. Then

##\displaystyle\frac{d}{dx}\,F(g(x))=F'(g(x))g'(x)##.

Thus,

$$\displaystyle\int_a^b{f(g(x))g'(x)}=F(g(x))\Bigg |_a^b=F(g(b))-F(g(a))

=F(B)-F(A)=F(u)\Bigg |_A^B=\displaystyle\int_A^B{\,f(u)du}$$

EXAMPLE 5 Evaluate the integral ##I=\displaystyle\int_0^8{\displaystyle\frac{\cos{\sqrt{x+1}}}{\sqrt{x+1}}\,dx}##.

Solution Let ##u=\sqrt{x+1}##. Then ##du=\displaystyle\frac{dx}{2\sqrt{x+1}}##. If ##x=0##, then ##u=1##; if ##x=8##, then ##u=3##. Thus

$$I=2\displaystyle\int_1^3{\cos{u}\,du}=2\sin{u\Bigg |_1^3}=2\sin{3}-2\sin{1}$$

Paradoxically, I understand the example 5, but can't deal with the theory, ie, the theorem 6. Try to focus the doubt about the theorem will be my attempt: why in the equality ##\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}##, at the left, the integrand is a composite function?.

Greetings

PD: Post without preview

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