Calculus Problem Set: Integration, Taylor Series, and Parametric Equations
Published:
1. Evaluate the integral
(a) $\int \frac{5x-3}{(x+1)(x-3)} dx$
(b) $\int e^{\tan x} \sec^2 x dx$
(c) $\int_0^{\pi/2} \frac{\cos x}{\sin x + 1} dx$
(d) $\int \frac{x+1}{x^2(x-1)} dx$
(e) $\int x^3 e^{4x} dx$
(f) $\int x(x-8)^{10} dt$
(g) $\int x^4 \ln 3x dx$
(h) $\int x^4(x+1)^{\frac{1}{3}} dx$
2. Using the substitution $t = \tan(\frac{x}{2})$, evaluate: $\int \frac{dx}{\sin x}$
3. Find all first and second order partial derivatives of $f(x, y) = x^2 + 6y^5 - 10xy$.
4. Construct the Taylor series expansion of
(a) $f(x) = \sin x$ about $x = \frac{\pi}{2}$.
(b) $f(x) = \ln x$ about $x = 1$.
(c) $f(x) = \frac{1}{x^2}$ about $x = 2$.
5. Find the length of the arc of the curve $y = (x-1)^{\frac{3}{2}}$ from $x = 1$ to $x = 5$.
6. In the Taylor series expansion of $f(x) = x^4$ centered at $a = 1$, what is the coefficient of $(x-1)^3$.
7. Consider the curve parameterized by $x = \frac{1}{3}t^3 + 3t^2 + 2$, $y = t^3 - t^2$,
(a) Find an equation for the line tangent to the curve when $t = 1$.
(b) Compute $\frac{d^2y}{dx^2}$ at $t = 1$.
8. Use the t-method method to evaluate the integral: $\int \frac{d\theta}{\dots}$
Animated Video Solution
The first half plays free, the full solution is in the app.
Step by Step Written Solution
Today we'll solve problem number five from this worksheet, which asks us to find the arc length of the curve defined by y equals x minus one to the power of three halves, from x equals one to x equals five.
Arc Length Calculation
First, let's recall the formula for the arc length of a curve defined as y equals f of x. The length L is the integral of the square root of one plus the derivative squared.
Before we can integrate, we need to find the first derivative of our function. Since y equals x minus one to the three halves, we apply the power rule.
Next, we need the square of the derivative to plug into our arc length formula. When we square three halves, we get nine fourths, and the square of x minus one to the one half is simply x minus one.
Now, let's substitute this into our integrand. We seek the integral of the square root of one plus nine fourths times the quantity x minus one.
Setting up the Integral
Let's simplify the expression inside the square root by distributing the nine fourths and combining it with the constant one.
One minus nine fourths is negative five fourths. So our integrand becomes the square root of nine fourths x minus five fourths.
To make the integration easier, let's factor out one fourth from under the radical. The square root of one fourth is one half, which we can move outside the integral.
Now we can evaluate this integral. This is a simple power rule integration where the inner function is linear, nine x minus five.
Integration Steps
The rest of this solution is on Solvi
8 more steps are locked. Watch the full animated, narrated solution for free.
Snap a photo, solve any question like this.
Watch the Rest for Free
Free to download · First solutions are on us
