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Compute the value of a function given by an integral formula

Compute f(2) for each of the following continuous functions f.

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

  1. Since

        \[ \int_0^x f(t) \, dt = x^2(1+x) = x^3 + x^2 \]

    we take the derivative of both sides to obtain

        \[ f(x) = 3x^2 + 2x. \]

    Therefore,

        \[ f(2) = 16. \]

  2. Since

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

    we take the derivative of both sides (see part (b) of this exercise for taking the derivative of an integral like this) to obtain

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

    Thus,

        \[ f(2) = f \left( \sqrt{2}^2 \right) = \frac{3 \sqrt{2}}{2} + 1. \]

  3. Here we have

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

    Here we evaluate the integral on the left,

        \[ \int_0^{f(x)} t^2 \, dt = \left. \frac{1}{3} t^3 \right|_0^{\f(x)} = \frac{1}{3} (f(x))^3. \]

    Therefore,

        \[ (f(x))^3 = x^3 + x^2 \implies f(x) = \left( 3(x^3 + x^2) \right)^{\frac{1}{3}}. \]

    Thus,

        \[ f(2) = (36)^{\frac{1}{3}}. \]

  4. Finally, since

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

    we take the derivative of both sides,

        \[ (3x^2 + 2x) f(x^3+x^2) = 1 \quad \implies \quad f(x^3+x^2) = \frac{1}{3x^2 + 2x}. \]

    Then, since

        \[ x^3 + x^2 = 2 \quad \iff \quad x = 1 \]

    (since the cubic polynomial x^3 + x^2 - 2 = 0 has only one real root at x = 1) we have

        \[ f(2) = \frac{1}{5}. \]

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