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To evaluate the integral integral (2x - x2)\,dx:

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Answer

x2−x33+Cx^2 - \frac{x^3}{3} + C

t+3t−13t3+Ct + \frac{3}{t} - \frac{1}{3t^3} + C

Step-by-step solution

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Here are the solutions for the integration problems:

18.
To evaluate the integral ∫(2x−x2) dx\int (2x - x^2)\,dx:

  1. Apply the sum/difference rule for integration.

    ∫(2x−x2) dx=∫2x dx−∫x2 dx\int (2x - x^2)\,dx = \int 2x\,dx - \int x^2\,dx

  2. Apply the power rule for integration ∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C.

    ∫2x dx=2⋅x1+11+1+C1=2⋅x22+C1=x2+C1\int 2x\,dx = 2 \cdot \frac{x^{1+1}}{1+1} + C_1 = 2 \cdot \frac{x^2}{2} + C_1 = x^2 + C_1 ∫x2 dx=x2+12+1+C2=x33+C2\int x^2\,dx = \frac{x^{2+1}}{2+1} + C_2 = \frac{x^3}{3} + C_2

  3. Combine the results and add the constant of integration CC.

    ∫(2x−x2) dx=x2−x33+C\int (2x - x^2)\,dx = x^2 - \frac{x^3}{3} + C

    The integral is x2−x33+C\boxed{x^2 - \frac{x^3}{3} + C}.

    19.
    To evaluate the integral ∫t4−3t2+1t4 dt\int \frac{t^4 - 3t^2 + 1}{t^4}\,dt:

  4. Divide each term in the numerator by the denominator to simplify the expression.

    t4−3t2+1t4=t4t4−3t2t4+1t4\frac{t^4 - 3t^2 + 1}{t^4} = \frac{t^4}{t^4} - \frac{3t^2}{t^4} + \frac{1}{t^4} =1−3t−2+t−4= 1 - 3t^{-2} + t^{-4}

  5. Integrate the simplified expression term by term.

    ∫(1−3t−2+t−4) dt=∫1 dt−∫3t−2 dt+∫t−4 dt\int \left(1 - 3t^{-2} + t^{-4}\right)\,dt = \int 1\,dt - \int 3t^{-2}\,dt + \int t^{-4}\,dt

  6. Apply the power rule for integration ∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C.

    ∫1 dt=t+C1\int 1\,dt = t + C_1 ∫3t−2 dt=3⋅t−2+1−2+1+C2=3⋅t−1−1+C2=−3t−1+C2\int 3t^{-2}\,dt = 3 \cdot \frac{t^{-2+1}}{-2+1} + C_2 = 3 \cdot \frac{t^{-1}}{-1} + C_2 = -3t^{-1} + C_2 ∫t−4 dt=t−4+1−4+1+C3=t−3−3+C3=−13t−3+C3\int t^{-4}\,dt = \frac{t^{-4+1}}{-4+1} + C_3 = \frac{t^{-3}}{-3} + C_3 = -\frac{1}{3}t^{-3} + C_3

  7. Combine the results and add the constant of integration CC.

    ∫t4−3t2+1t4 dt=t−(−3t−1)+(−13t−3)+C\int \frac{t^4 - 3t^2 + 1}{t^4}\,dt = t - (-3t^{-1}) + \left(-\frac{1}{3}t^{-3}\right) + C =t+3t−1−13t−3+C= t + 3t^{-1} - \frac{1}{3}t^{-3} + C

  8. Express with positive exponents.

    =t+3t−13t3+C= t + \frac{3}{t} - \frac{1}{3t^3} + C

    The integral is t+3t−13t3+C\boxed{t + \frac{3}{t} - \frac{1}{3t^3} + C}.

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