Integration by subtitution

Question

Can someone explain me how to find the value of

$$L = \int_1^2 \sqrt{1+9x} \,\mathrm{d}x$$

I do not know how to approach it after having $z = 1+9x$ and $\mathrm{d}x = \mathrm{d}z/9$.

Answer 1

Your substitution gives

$$L=\frac{1}{9}\int_{10}^{19}\sqrt{z}\,\mathrm{d}z=\frac{1}{9}\left[\frac23z^{3/2}\right]_{10}^{19}=\frac{2}{27}\left(19^{3/2}-10^{3/2}\right)$$

Notice that for $a\neq-1$, a primitive of $z^a$ is

$$\frac{z^{a+1}}{a+1}$$

And $\sqrt{z}=z^ {1/2}$.

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When doing integration by substitution of

$$I=\int_a^b f(x)\,\mathrm{d}x$$

With substitution $x=g(z)$, with a suitable $g$ (differentiable and bijective), you write $\mathrm{d}x=g'(z)\,\mathrm{d}z$ and

$$I=\int_{g^{-1}(a)}^{g^{-1}(b)} f(g(z)) \,g'(z)\,\mathrm{d}z$$

Actually, you simply replace $x$ with $g(z)$ in the integrand, you replace $\mathrm{d}x$ with $g'(z)\,\mathrm{d}z$, and you also replace the bounds $(a,b)$ with $(z_1,z_2)$. From $a=g(z_1)$ you get $z_1=g^{-1}(a)$ and likewise $z_2=g^{-1}(b)$.

Here, $1+9x=z$, so $x=g(z)=\frac{z-1}{9}$, and $a=1,b=2$ so $z_1=1+9a=10$ and $z_2=1+9b=19$.