Page 117 of 173

Learn how to integrate an absolute value
Learn how to integrate an absolute value

Compute the derivative f'(x) for the function f(x) defined as

\[ f(x) = \int_{x^3}^{x^2} \frac{t^6}{1+t^4} \, dt \]

without evaluating the integral directly.

Let

\[ A(x) = \int_0^x \frac{t^6}{1+t^4} \, dt. \]

Then, by the fundamental theorem of calculus,

\[ A'(x) = \frac{x^6}{1+x^4}. \]

So, using the linearity of the integral with respect to the interval of integration we write,

\begin{align*} f(x) &= \int_{x^3}^{x^2} \frac{t^6}{1+t^4} \, dt \\ &= \int_0^{x^2} \frac{t^6}{1+t^4} \, dt - \int_0^{x^3} \frac{t^6}{1+t^4} \, dt \\ &= A(x^2) - A(x^3). \end{align*}

Now, differentiating and using the chain rule we have

\begin{align*} f'(x) &= (A(x^2))' - (A(x^3))' \\ &= (2x) A'(x^2) - (3x^2) A'(x^3) \\ &= (2x) \frac{x^{12}}{1+x^8} - (3x^2) \frac{x^{18}}{1+x^{12}} \\ &= \frac{2x^{13}}{1+x^8} - \frac{3x^{20}}{1+x^{12}}. \end{align*}

Find the derivative f'(x) for each of the following (without attempting to evaluate the integrals)

  1. \displaystyle{f(x) = \int_0^x (1+t^2)^{-3} \, dt},
  2. \displaystyle{f(x) = \int_0^{x^2} (1+t^2)^{-3} \, dt},
  3. \displaystyle{f(x) = \int_{x^3}^{x^2} (1+t^2)^{-3} \, dt}.

For all parts of the problem let

\[ A(x) = \int_0^x (1+t^2)^{-3} \, dt. \]

From the fundamental theorem of calculus we then know

\[ A'(x) = (1+x^2)^{-3}. \]

  1. Now, f(x) = A(x) so

    \[ f'(x) = A'(x) = (1+x^2)^{-3}. \]

  2. In this case f(x) = A(x^2) so

    \[ f'(x) = (A(x^2))' = (2x)A'(x^2) = (2x)(1+x^4)^{-3}. \]

  3. Now, we rewrite the integral,

    \[ f(x) = \int_{x^3}^{x^2} (1+t^2)^{-3} \, dt = \int_0^{x^2} (1+t^2)^{-3} - \int_0^{x^3} (1+t^2)^{-3} \, dt. \]

    Therefore,

    \[ f(x) = A(x^2) - A(x^3) \quad \implies \quad f'(x) = (A(x^2))' - (A(x^3))'. \]

    From above we know

    \begin{align*} (A(x^2))' = (2x) A'(x^2) &= (2x)(1+x^4)^{-3} \\ (A(x^3))' = (3x^2) A'(x^3) &= (3x^2)(1+x^6)^{-3}. \end{align*}

    Therefore,

    \[ f'(x) = (2x)(1+x^4)^{-3} - 3x^2(1+x^6)^{-3}. \]

Let g(x) be an everywhere continuous function with

\[ g(1) = 5 \qquad \text{and} \qquad \int_0^1 g(t) \, dt = 2. \]

Define another function f(x) by

\[ f(x) = \frac{1}{2} \int_0^x (x-t)^2 g(t) \, dt. \]

Prove that

\[ f'(x) = x \int_0^x g(t) \, dt - \int_0^x t g(t) \, dt. \]

Also, compute the values, f''(1) and f'''(1).

Proof. We start with the equation for f(x) and use properties of the integral and the fundamental theorem of calculus to rearrange,

\begin{align*} f(x) &= \frac{1}{2} \int_0^x (x-t)^2 g(t) \, dt \\ &= \frac{1}{2} \int_0^x (x^2 g(t) - 2xt g(t) + t^2 g(t)) \,dt \\ &= \frac{1}{2} \int_0^x x^2 g(t) \, dt - \int_0^x xt g(t) \, dt + \frac{1}{2} \int_0^x t^2 g(t) \, dt &(\text{linearity of integral})\\ &= \frac{x^2}{2} \int_0^x g(t) \, dt - x \int_0^x t g(t) \, dt + \frac{1}{2} \int_0^x t^2 g(t). \end{align*}

We can pull the x out of the integral since it does not depend on t (which is what we are taking the integral over). Then, we want to take the derivative with respect to x; however, we need to be careful and use the product rule. Looking closely at the first term we have

\begin{align*} \frac{d}{dx}\left( \frac{x^2}{2} \int_0^x g(t) \, dt\right) &= \left( \frac{x^2}{2} \right)' \cdot \int_0^x g(t) \, dt + \frac{x^2}{2} \left( \int_0^x g(t) \, dt \right)' \\ &= x \int_0^x g(t) \, dt + \frac{x^2}{2} g(x). \end{align*}

Where we used the fundamental theorem to obtain \left( \int_0^x g(t) \, dt \right)' = g(x). So, using this method to take the derivative for f(x) we have

\begin{align*} f'(x) &= x \int_0^x g(t) \, dt + \frac{x^2}{2} g(x) - \int_0^x t g(t) \, dt - x^2 g(x) + \frac{x^2}{2} g(x) \\ &= x \int_0^x g(t) \, dt - \int_0^x t g(t) \, dt. \qquad \blacksquare \end{align*}

Now, to evaluate the second and third derivatives at 1 as requested, we first get expressions for f''(x) and f'''(x).

\begin{align*} f''(x) &= \int_0^x g(t) \, dt + x g(x) - x g(x) = \int_0^x g(t) \, dt. \\ f'''(x) &= \left( \int_0^x g(t) \, dt \right)' = g(x). \end{align*}

Therefore,

\begin{align*} f''(1) &= \int_0^1 g(t) \, dt = 2 \\ f'''(1) &= g(1) = 5. \end{align*}

Let

\[ f(x) = 3 + \int_0^x \frac{1+ \sin t}{2 + t^2} \, dt \qquad \text{for all } x \in \mathbb{R}. \]

Find a quadratic polynomial p(x) = a + bx + cx^2 such that

\[ p(0) = f(0), \quad p'(0) = f'(0), \quad \text{and} \quad p''(0) = f''(0). \]

Do not attempt to evaluate the integral.

First,

\[ f(0) = 3 + \int_0^0 \frac{1+\sin t}{2 + t^2} \, dt = 3. \]

Hence,

\[ p(0) = 3 \quad \implies \quad a = 3. \]

Next, by the fundamental theorem of calculus,

\[ f'(x) = \frac{1+\sin x}{2 + x^2} \quad \implies \quad f'(0) = \frac{1}{2}. \]

Then, evaluate the derivative of the polynomial,

\[ p'(x) = b + 2cx \quad \implies \quad p'(0) = \frac{1}{2} = b. \]

To get the final constant we take the derivative of f'(x),

\begin{align*} f''(x) &= \frac{ (\cos x)(2+x^2) - (1+\sin x)(2x)}{(2+x^2)^2} \\ \implies \quad f''(0) &= \frac{1}{2}. \end{align*}

Therefore,

\[ p''(x) = 2c \quad \implies \quad p''(0) = \frac{1}{2} = 2c \quad \implies \quad c = \frac{1}{4}. \]

Hence, we have

\[ p(x) = 3 + \frac{1}{2} x + \frac{1}{4} x^2. \]

Find a function f and a constant c such that

\[ \int_0^x f(t) \, dt = \int_x^1 t^2 f(t) \, dt + \frac{x^{16}}{8} + \frac{x^{18}}{9} + c \qquad \text{for all } x \in \mathbb{R}. \]

Let P(x) = \int_0^x t^2 f(t) \, dt. Then, taking derivatives, and using the fundamental theorem of calculus we have

\begin{align*} && f(x) &= (P(1) - P(x))' + 2x^{15} + 2x^{17} \\ \implies && f(x) &= -x^2 f(x) + 2x^{15} + 2x^{17} \\ \implies && f(x)(x^2+1) &= 2x^{15} (x^2+1) \\ \implies && f(x) &= 2x^{15}. \end{align*}

Then, to find the value of the constant, let x = 0 and evaluate

\begin{align*} &&0 &= \int_0^1 t^2 f(t) \, dt + c \\ \implies && c &= -\int_0^1 t^2 f(t) \, dt \\ \implies && c &= -\int_0^1 2t^{17} \, dt \\ \implies && c &= - \frac{1}{9} t^{18} \bigr \rvert_0^1 \\ \implies && c &= -\frac{1}{9}. \end{align*}

Find a function f and a constant c such that

\[ \int_c^x t f(t) \, dt = \sin x - x \cos x - \frac{1}{2} x^2 \qquad \text{for all } x \in \mathbb{R}. \]

Let f(t) = \sin t - 1 and c = 0. Then,

\begin{align*} \int_c^x t f(t) \, dt &= \int_0^x (t \sin t - t) \, dt \\ &= \left. \left( \sin t - t \cos t - \frac{1}{2} t^2 \right) \right|_0^x \\ &= \sin x - x \cos x - \frac{1}{2} x^2. \end{align*}

Find a function f and a constant c such that

\[ \int_c^x f(t) \, dt = \cos x - \frac{1}{2} \qquad \text{for all } x \in \mathbb{R}. \]

Let f(t) = -\sin t and c = \frac{\pi}{3}. Then,

\begin{align*} \int_c^x f(t) \, dt &= \int_{\frac{\pi}{3}}^x (- \sin t) \, dt \\ &= \cos t \bigr \rvert_{\frac{\pi}{3}}^x \\ &= \cos x - \cos \left( \frac{\pi}{3} \right) \\ &= \cos x - \frac{1}{2}. \end{align*}

Let f be an everywhere continuous function satisfying

\[ \int_0^x f(t) \, dt = -\frac{1}{2} + x^2 + x \sin (2x) + \frac{1}{2} \cos (2x) \]

for all x. Find the values of

\[ f \left( \frac{\pi}{4} \right) \qquad \text{and} \qquad f' \left( \frac{\pi}{4} \right). \]

Let

\[ A(x) = \int_0^x f(t) \, dt = -\frac{1}{2} + x^2 + x \sin (2x) + \frac{1}{2} \cos (2x). \]

By the first fundamental theorem of calculus we know the derivative A'(x) exists (since f is continuous by assumption) and we have

\[ A'(x) = f(x). \]

Therefore, taking the derivative of A(x) we have,

\[ f(x) = A'(x) = 2x + \sin (2x) + 2x \cos (2x) - \sin (2x) = 2x + 2x \cos (2x) \]

Then, evaluating at \frac{\pi}{4},

\begin{align*} f \left( \frac{\pi}{4} \right) &= A' \left( \frac{\pi}{4} \right) \\ &= \frac{\pi}{2} + \frac{\pi}{2} \cos \left( \frac{\pi}{2} \right) \\ &= \frac{\pi}{2}. \end{align*}

Then, taking the derivative of f(x) we have,

\[ f(x) = 2x + 2x \cos (2x) \quad \implies \quad f'(x) = 2 + 2 \cos(2x) - 4x \sin (2x). \]

Therefore,

\[ f' \left( \frac{\pi}{4} \right) = 2 + 2 \cos \left( \frac{\pi}{2} \right) - \pi \sin \left( \frac{\pi}{2} \right) = 2 - \pi. \]

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