Integrale berechnen

Exercise 1

Bestimme den Wert des Integrals.

  1. ∫02(3x2−4x+5) dx\int_{0}^{2} (3x^2 - 4x + 5) \, dx

  2. ∫131x2 dx\int_{1}^{3} \frac{1}{x^2} \, dx

  3. ∫01(ex+2) dx\int_{0}^{1} (e^x + 2) \, dx

  4. ∫0πsin⁡(x) dx\int_{0}^{\pi} \sin(x) \, dx

  5. ∫19x dx\int_{1}^{9} \sqrt{x} \, dx

  6. ∫122x3 dx\int_{1}^{2} \frac{2}{x^3} \, dx

  7. ∫08x3 dx\int_{0}^{8} \sqrt[3]{x} \, dx

  8. ∫1e4x dx\int_{1}^{e} \frac{4}{x} \, dx

  9. ∫01(10x4−2x) dx\int_{0}^{1} (10x^4 - 2x) \, dx

  10. ∫121x4 dx\int_{1}^{2} \frac{1}{x^4} \, dx

  11. ∫0π/22cos⁡(x) dx\int_{0}^{\pi/2} 2\cos(x) \, dx

  12. ∫493x dx\int_{4}^{9} \frac{3}{\sqrt{x}} \, dx

  13. ∫01(ex+x2) dx\int_{0}^{1} (e^x + x^2) \, dx

  14. ∫0π(cos⁡(x)+1) dx\int_{0}^{\pi} (\cos(x) + 1) \, dx

  15. ∫01x4 dx\int_{0}^{1} \sqrt[4]{x} \, dx

  16. ∫12(3x2+1) dx\int_{1}^{2} (\frac{3}{x^2} + 1) \, dx

  17. ∫01(x+x) dx\int_{0}^{1} (x + \sqrt{x}) \, dx

  18. ∫0π/2(sin⁡(x)+1) dx\int_{0}^{\pi/2} (\sin(x) + 1) \, dx

  19. ∫181x23 dx\int_{1}^{8} \frac{1}{\sqrt[3]{x^2}} \, dx

  20. ∫02(x3−3x2) dx\int_{0}^{2} (x^3 - 3x^2) \, dx

Solution
  1. [x3−2x2+5x]02=(8−8+10)−0=10[x^3 - 2x^2 + 5x]_0^2 = (8 - 8 + 10) - 0 = \mathbf{10}
  2. [−x−1]13=[−1x]13=−13−(−1)=23[-x^{-1}]_1^3 = [-\frac{1}{x}]_1^3 = -\frac{1}{3} - (-1) = \mathbf{\frac{2}{3}}
  3. [ex+2x]01=(e+2)−(e0+0)=e+2−1=e+1[e^x + 2x]_0^1 = (e + 2) - (e^0 + 0) = e + 2 - 1 = \mathbf{e + 1}
  4. [−cos⁡(x)]0π=−(−1)−(−1)=1+1=2[-\cos(x)]_0^{\pi} = -(-1) - (-1) = 1 + 1 = \mathbf{2}
  5. [23x3/2]19=23(27−1)=23⋅26=523[\frac{2}{3}x^{3/2}]_1^9 = \frac{2}{3}(27 - 1) = \frac{2}{3} \cdot 26 = \mathbf{\frac{52}{3}} (ca. 17.33)
  6. [−x−2]12=[−1x2]12=−14−(−1)=34[-x^{-2}]_1^2 = [-\frac{1}{x^2}]_1^2 = -\frac{1}{4} - (-1) = \mathbf{\frac{3}{4}}
  7. [34x4/3]08=34(16−0)=12[\frac{3}{4}x^{4/3}]_0^8 = \frac{3}{4}(16 - 0) = \mathbf{12}
  8. [4ln⁡(x)]1e=4ln⁡(e)−4ln⁡(1)=4(1)−0=4[4\ln(x)]_1^e = 4\ln(e) - 4\ln(1) = 4(1) - 0 = \mathbf{4}
  9. [2x5−x2]01=(2−1)−0=1[2x^5 - x^2]_0^1 = (2 - 1) - 0 = \mathbf{1}
  10. [−13x−3]12=[−13x3]12=−124−(−13)=724[-\frac{1}{3}x^{-3}]_1^2 = [-\frac{1}{3x^3}]_1^2 = -\frac{1}{24} - (-\frac{1}{3}) = \frac{7}{24}
  11. [2sin⁡(x)]0π/2=2(1)−0=2[2\sin(x)]_0^{\pi/2} = 2(1) - 0 = \mathbf{2}
  12. [6x]49=6(3−2)=6[6\sqrt{x}]_4^9 = 6(3 - 2) = \mathbf{6}
  13. [ex+13x3]01=(e+13)−(1+0)=e−23[e^x + \frac{1}{3}x^3]_0^1 = (e + \frac{1}{3}) - (1 + 0) = \mathbf{e - \frac{2}{3}}
  14. [sin⁡(x)+x]0π=(0+π)−(0+0)=π[\sin(x) + x]_0^{\pi} = (0 + \pi) - (0 + 0) = \mathbf{\pi}
  15. [45x5/4]01=45[\frac{4}{5}x^{5/4}]_0^1 = \mathbf{\frac{4}{5}} (oder 0.8)
  16. [−3x−1+x]12=[−3x+x]12=(−1,5+2)−(−3+1)=0,5−(−2)=2,5[-3x^{-1} + x]_1^2 = [-\frac{3}{x} + x]_1^2 = (-1,5 + 2) - (-3 + 1) = 0,5 - (-2) = \mathbf{2,5}
  17. [12x2+23x3/2]01=12+23=76[\frac{1}{2}x^2 + \frac{2}{3}x^{3/2}]_0^1 = \frac{1}{2} + \frac{2}{3} = \mathbf{\frac{7}{6}}
  18. [−cos⁡(x)+x]0π/2=(0+π2)−(−1+0)=π2+1[-\cos(x) + x]_0^{\pi/2} = (0 + \frac{\pi}{2}) - (-1 + 0) = \mathbf{\frac{\pi}{2} + 1}
  19. [3x1/3]18=3(2−1)=3[3x^{1/3}]_1^8 = 3(2 - 1) = \mathbf{3}
  20. [14x4−x3]02=(164−8)−0=4−8=−4[\frac{1}{4}x^4 - x^3]_0^2 = (\frac{16}{4} - 8) - 0 = 4 - 8 = \mathbf{-4}
Exercise 2

Überprüfe einige der Aufgaben mit dem Taschenrechmner!