Note that the type of the first function was considered in Exercise 1 on Long Day this week and the types of the other two in Exercise 2 on Long Day this week.
hint
For the second function, use the rewriting $\displaystyle{\frac{1}{x^3}=x^{-3}}\,.$
Determine an antiderivative for the function $\,x\cos(x)\,,$ and then do a check to ensure that your result is correct.
hint
Use Theorem 23.7 and see Example 23.8.
hint
It will be smart to choose $f(x)=\cos(x)$ and $g(x)=x\,.$
hint
“Checking” simply means that you differentiate your resulting antiderivative to ensure that its derivative indeed is the function you started out with.
answer
$$\cos(x)+x\sin(x)$$
Since $(\cos(x)+x\sin(x))’=-\sin(x)+\sin(x)+x\cos(x)=x\cos(x)\,,$ then the found antiderivative is correct.
B
Determine the indefinite integral $\displaystyle{\,\int{t\e^t \,\mathrm dt}\,},$ and then check your result.
hint
Use Theorem 23.7 by setting $f(t)=\e^t$ and $g(t)=t\,.$
hint
Don’t forget the constant.
answer
$$t\e^t-\e^t+k\,, k\in \reel$$
C
Determine an antiderivative of the function $\,x^2\ln(x)\,,\,\,x>0\,.$
hint
Use Theorem 23.7 by setting $f(x)=x^2$ and $g(x)=\ln(x)\,.$
where $a$ and $b$ are parameter interval end points for your parametrization of $K$.
hint
$f(\mathbf r(u))$ is the restriction of the function to the curve $K$ that you have found the parametrization $\mathbf r(u)$ for. The function would be $f(x)=x^2\,.$
hint
$$f(\mathbf r(u))=u^2$$
hint
The integrand becomes $\displaystyle{u\sqrt{u^2+1}}\,.$ Find an antiderivative of the integrand by the method of substitution. For this, choose $g(x)=x^2+1$ as an inner function and $f(x)=\sqrt x$ as an outer function.
A solid model of The Gherkin appears when we rotate region $\,A\,$ about the $\,z$-axis by an angle of $\,2\pi\,.$ Determine the volume of this model.
hint
Can you construct a volume formula? Next week we will develop such a formula as a tripple integration, but for this example today a volume formula can be constructed in a simpler manner.
hint
Think of the volume of the model as the total of many, many horizontal slices. Each slice is a circular disc.
Essentially this is a summation of infinitely many (thus, an integral) circular horizontal “slices” through the model, each with an area of $\pi f(z)^2$, where $f(z)$ constitutes the radius of the “slice”.
answer
$$\frac 94 \pi$$
B
Determine the area of region $\,A\,.$
hint
Quite some mathematical acrobatics is needed to find an indefinite integral of $\,f\,.$ Choose the inner function as $\,\displaystyle{g(z)=\frac 12 z -\frac 12\,}$ and the outer function as $\,\displaystyle{h(z)=2\sqrt{1-z^2}}\,,$ and compute $\,h(g(z))g’(z)\,.$