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Evaluate the integral: integral (2x3-5x2+4x-1)\,dx

Mathematics

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Answer

x42−5x33+2x2−x+C\frac{x^4}{2} - \frac{5x^3}{3} + 2x^2 - x + C

Step-by-step solution

5 steps

To evaluate the integral ∫(2x3−5x2+4x−1) dx\int (2x^3-5x^2+4x-1)\,dx, we will integrate each term separately using the power rule for integration, which states ∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C.

  1. Integrate the first term, 2x32x^3.

    ∫2x3 dx=2⋅x3+13+1=2⋅x44=x42\int 2x^3\,dx = 2 \cdot \frac{x^{3+1}}{3+1} = 2 \cdot \frac{x^4}{4} = \frac{x^4}{2}
  2. Integrate the second term, −5x2-5x^2.

    ∫−5x2 dx=−5⋅x2+12+1=−5⋅x33=−5x33\int -5x^2\,dx = -5 \cdot \frac{x^{2+1}}{2+1} = -5 \cdot \frac{x^3}{3} = -\frac{5x^3}{3}
  3. Integrate the third term, $4x$.

    ∫4x dx=4⋅x1+11+1=4⋅x22=2x2\int 4x\,dx = 4 \cdot \frac{x^{1+1}}{1+1} = 4 \cdot \frac{x^2}{2} = 2x^2
  4. Integrate the fourth term, −1-1.

    ∫−1 dx=−x\int -1\,dx = -x
  5. Combine all the integrated terms and add the constant of integration, CC.

    ∫(2x3−5x2+4x−1) dx=x42−5x33+2x2−x+C\int (2x^3-5x^2+4x-1)\,dx = \frac{x^4}{2} - \frac{5x^3}{3} + 2x^2 - x + C

    The evaluated integral is:

    x42−5x33+2x2−x+C\frac{x^4}{2} - \frac{5x^3}{3} + 2x^2 - x + C
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