10.2 Quiz 1 solutions

Q1.1 The answer is D. To evaluate the integral we substitute $u=\sqrt{1+x^{2}}$, so that $\frac{du}{dx}=\frac{1}{2}\cdot 2x\cdot(1+x^{2})^{-\frac{1}{2}}=\frac{x}{\sqrt{1+x^{2}}}$. The bounds change as follows: $u=1$ when $x=0$, and $u=\sqrt{2}$ when $x=1$. So $I=\int_{1}^{\sqrt{2}}e^{u}\,du=[e^{u}]_{1}^{\sqrt{2}}=e^{\sqrt{2}}-e$.

Q1.2. The answer is B. To evaluate the integral we substitute $t=\tan\frac{x}{2}$, see slide 1.32 in the notes. Then $\frac{dx}{dt}=\frac{2}{1+t^{2}}$, $\sin x=\frac{2t}{1+t^{2}}$, $\cos x=\frac{1-t^{2}}{1+t^{2}}$ and the limits of integration change as: $t=1$ when $x=\frac{\pi}{2}$, $t=-1$ when $x=-\frac{\pi}{2}$. So

which we can solve by the further substitution $u=\frac{t+1}{2}$, so that $\frac{dt}{du}=2$ and the limits of integration change as: $u=0$ when $t=-1$, $u=1$ when $t=1$, therefore:

Q1.3 The answer is A. We substitute $x=\sin t$, so the bounds change as: $t=0$ when $x=0$ and $t=\frac{\pi}{4}$ when $x=\frac{1}{\sqrt{2}}$. Then $\frac{dx}{dt}=\cos t$ and $\sqrt{1-x^{2}}=\sqrt{1-\sin^{2}t}=|\cos t|$. We have $\cos t>0$ for $0\leq t\leq\frac{\pi}{4}$, so $(1-x^{2})^{\frac{3}{2}}=\cos^{3}t$. Thus $K=\int_{0}^{\frac{\pi}{4}}\frac{\cos t\,dt}{\cos^{3}t}=\int_{0}^{\frac{\pi}{4}}\sec^{2}t\,dt=\left[\tan t\right]_{0}^{\frac{\pi}{4}}=1$.

Q1.4 The answer is B. The other commands are respectively:

(A) $\tan^{-1}x\leq\log\frac{3}{y}$ or $\tan^{-1}x\leq\sinh z$;

(C) $\tan^{-1}x\geq\log\frac{3}{y}$ and $\tan^{-1}x\geq\sinh z$;

(D) $\tan^{-1}x\geq\log\frac{3}{y}$ or $\tan^{-1}x\geq\sinh z$;

(E) $\cot x\geq\log\frac{3}{y}$ and $\cot x\geq\sinh z$.

(In (E), don’t be misled by the slightly confusing notation for trigonometric functions: $\tan^{2}x=(\tan x)^{2}$, but $\tan^{-1}x\neq(\tan x)^{-1}$.)

Q1.5 The answer is 53. The following code, for example, will give you this answer.

x <- seq(1:100)

y <- sin(3*x*pi/44)

condition <- y > 0.1

length(y[condition])