Back to Section Page Back to Problems Page

Practice Problems for 4/29

  1. Suppose \(F(x)=\int_0^x \frac{2}{t^3+1} dt\). Explain why \(F\) is differentiable at \(x=1\), then find the value of \(F'(1)\).
    2. The function \(\frac{2}{x^3+1}\) is continuous at all \(x\neq -1\). It is therefore continuous on [0,b], for any \(b>1\), so by the Fundamental Theorem of Calculus Part I, \(F\) is differentiable at \(x=1\) and \(F'(1)=\frac{2}{1^3+1}=1\).
  2. Compute \(\frac{d}{dx}\int_x^{\sin x} 3t^2\, dt\).
    3. In order to apply FTC Part I, one of the integral bounds must be a constant. The first step is therefore to write this single integral as the sum of two integrals using the Additivity property:
    \[ \int_x^{\sin x} 3t^2\, dt=\int_x^{0} 3t^2\, dt + \int_0^{\sin x} 3t^2\, dt. \] Note that any constant could have been used in place of zero. Now, we can apply FTC I to each integral since the integrand, \(3x^2\), is everywhere continuous. \[ \frac{d}{dx} \int_x^{0} 3t^2\, dt=\frac{d}{dx}-\int_0^{x} 3t^2\, dt=-3x^2, \] and \[ \frac{d}{dx} \int_0^{\sin x}3t^2\, dt=3\sin^2 x \frac{d}{dx}\sin x=3\sin^2 x \cos x, \] where we have applied the chain rule. Hence, \[ \frac{d}{dx}\int_x^{\sin x} 3t^2\, dt=-3x^2 + 3\sin^2 x \cos x. \]
  3. Which of the following can be evaluated using the Fundamental Theorem of Calculus Part 2?
    1. \( \int_1^5 \frac{x-3}{x^2+1}\,dx\)

    2. \( \int_1^5 \frac{x-3}{x^2-4}\, dx\)

    3. \(\int_{-1}^1 \frac{1}{x}\, dx\)

    4. \(\int_{1/2}^1 \frac{1}{x}\, dx\)

    5. \( \int_{-1}^1 \sqrt{1-x^2}\,dx\)

    4. FTC II requires \(f(x)\) to be continuous on all of \([a,b]\). It therefore only applies to the definite integrals a, d & e. It does not apply to part b because \(f(x)=\frac{x-3}{x^2-4}\) has an infinite discontinuity at \(x=2\), which is in the interval [1,5]. Similarly, FTC II does not apply to part c because \(f(x)=\frac{1}{x}\) has an infinite discontinuity at \(x=0\), which is contained in [-1,1].
  4. Evaluate \( \int_0^1 4x^3-9x^2+2x-3\, dx\)
    5. Since \(f(x)=4x^3-9x^2+2x-3\) is continuous on [0,1], we can apply FTC II. \[ \int_0^1 4x^3-9x^2+2x-3\, dx=x^4-3x^3+x^2-3x |_0^1=(1-3+1-3)-(0-0+0-0)=-4. \]