Use the crossing before integrating an absolute value · source reference

Printed question

(a) Use $u=e^x+2$ to find the exact value of $\int_0^{\ln2}\frac{e^{3x}}{e^x+2}dx$. [4]

(b) Sketch $y=e^x-2$ for $0\le x\le2$. State the exact $x$ where it crosses the $x$-axis. [2]

Hence find the exact value of $\ln k$ such that

$$\int_0^2|e^x-2|dx=\int_e^{e^{\sqrt k}}\frac1{x\ln x}dx.$$

[5]

[2011/TPJC/Prelim/I/9]

Printed working

(a)

$u=e^x+2$, $du/dx=e^x$; when $x=0,u=e^0+2=3$; when $x=\ln2,u=e^{\ln2}+2=4$.

$$\int_0^{\ln2}\frac{e^{3x}}{e^x+2}dx=\int_3^4\frac{e^{3x}}{e^x+2}\frac1{e^x}du=\int_3^4\frac{(u-2)^2}{u}du=\int_3^4\frac{u^2-4u+4}{u}du=\int_3^4(u-4+4/u)du=[u^2/2-4u+4\ln u]_3^4=4\ln(4/3)-\frac12.$$

(b)

$y=e^x-2$ crosses the $x$-axis at $x=\ln2$; endpoints $(0,-1)$ and $(2,e^2-2)$.

$$\begin{aligned}\int_0^{\ln2}-(e^x-2)dx+\int_{\ln2}^2(e^x-2)dx&=\int_e^{e^{\sqrt k}}\frac{1/x}{\ln x}dx\\-[e^x-2x]_0^{\ln2}+[e^x-2x]_{\ln2}^2&=[\ln|\ln x|]_e^{e^{\sqrt k}}\\-(2-2\ln2-1)+(e^2-4-2+2\ln2)&=\ln\sqrt k=\frac12\ln k.\end{aligned}$$ $$\ln k=2(e^2+4\ln2-7).$$

Teacher notes

Substitution: $\int f(x)dx=\int f(x)(dx/du)du=\int g(u)(dx/du)du$. For a definite integral change limits from $x=\cdots$ to $u=\cdots$, here $\int_{x=0}^{x=\ln2}f(x)dx=\int_{u=3}^{u=4}g(u)du$. Express the integrand entirely in $u$.

Use $\int f'/f=\ln|f|+c$. The modulus sign in $\ln|\ln x|$ is highlighted.

Printed answer key

(a) $4\ln(4/3)-1/2$; (b) the native exponential graph below, $x=\ln2$; $\ln k=2(e^2+4\ln2-7)$.

Source diagrams in native form