8. Evaluate $\int_{0}^{(\sqrt{2}-1) / 2} \frac{\mathrm{d} x}{(2 x+1) \sqrt{x^{2}+x}}$.
Solution: Let $u=\sqrt{x^{2}+x}$. Then $d u=\frac{2 x+1}{2 \sqrt{x^{2}+x}} d x$. So the integral becomes $2 \int \frac{d u}{\left(4 x^{2}+4 x+1\right)}$, or $2 \int \frac{d u}{4 u^{2}+1}$. This is $\tan ^{-1}(2 u)$, yielding a final answer of $\tan ^{-1}\left(2 \sqrt{x^{2}+x}\right)+C$ for the indefinite integral. The definite integral becomes $\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{\pi}{4}$.
9. Suppose $f$ is a differentiable real function such that $f(x)+f^{\prime}(x) \leq 1$ for all $x$, and $f(0)=0$. What is the largest possible value of $f(1)$ ? (Hint: consider the function $e^{x} f(x)$.)

Solution: $1-1 / e$ Let $g(x)=e^{x} f(x)$; then $g^{\prime}(x)=e^{x}\left(f(x)+f^{\prime}(x)\right) \leq e^{x}$. Integrating from 0 to 1 , we have $g(1)-g(0)=\int_{0}^{1} g^{\prime}(x) d x \leq \int_{0}^{1} e^{x} d x=e-1$. But $g(1)-g(0)=e \cdot f(1)$, so we get $f(1) \leq(e-1) / e$. This maximum is attained if we actually have $g^{\prime}(x)=e^{x}$ everywhere; this entails the requirement $f(x)+f^{\prime}(x)=1$, which is met by $f(x)=1-e^{-x}$.
10. A continuous real function $f$ satisfies the identity $f(2 x)=3 f(x)$ for all $x$. If $\int_{0}^{1} f(x) d x=1$, what is $\int_{1}^{2} f(x) d x$ ?

Solution: 5 Let $S=\int_{1}^{2} f(x) d x$. By setting $u=2 x$, we see that $\int_{1 / 2}^{1} f(x) d x=$ $\int_{1 / 2}^{1} f(2 x) / 3 d x=\int_{1}^{2} f(u) / 6 d u=S / 6$. Similarly, $\int_{1 / 4}^{1 / 2} f(x) d x=S / 36$, and in general $\int_{1 / 2^{n}}^{1 / 2^{n-1}} f(x) d x=S / 6^{n}$. Adding finitely many of these, we have $\int_{1 / 2^{n}}^{1} f(x) d x=S / 6+S / 36+$ $\cdots+S / 6^{n}=S \cdot\left(1-1 / 6^{n}\right) / 5$. Taking the limit as $n \rightarrow \infty$, we have $\int_{0}^{1} f(x) d x=S / 5$. Thus $S=5$, the answer.