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Evaluate the integral from 1 to 2 of (3x2 - 1)dx

Mathematics

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: Evaluate ∫12(3x2−1) dx\int_{1}^{2} (3x^2 - 1)\,dx

  1. Find the indefinite integral of the function.

    ∫(3x2−1) dx=3x2+12+1−1x0+10+1+C=x3−x+C\int (3x^2 - 1)\,dx = \frac{3x^{2+1}}{2+1} - \frac{1x^{0+1}}{0+1} + C = x^3 - x + C
  2. Evaluate the definite integral using the Fundamental Theorem of Calculus.

    [x3−x]12=(23−2)−(13−1)\left[x^3 - x\right]_{1}^{2} = (2^3 - 2) - (1^3 - 1)
  3. Calculate the values.

    (8−2)−(1−1)=6−0=6(8 - 2) - (1 - 1) = 6 - 0 = 6

    The correct option is A.

    A. 6\boxed{\text{A. } 6}

    : In the diagram below, ABC is an isosceles triangle, ∣AC∣=12cm|AC|= 12\text{cm}, ∣AB∣=∣BC∣|AB| = |BC| and ∠ABC=120∘\angle ABC = 120^\circ. Find ∣AB∣|AB|.

  4. Find the base angles of the isosceles triangle.

    Since ∣AB∣=∣BC∣|AB| = |BC|, the angles opposite these sides are equal: ∠BAC=∠BCA\angle BAC = \angle BCA.
    The sum of angles in a triangle is 180∘180^\circ.

    ∠BAC+∠BCA+∠ABC=180∘\angle BAC + \angle BCA + \angle ABC = 180^\circ 2×∠BAC+120∘=180∘2 \times \angle BAC + 120^\circ = 180^\circ 2×∠BAC=60∘2 \times \angle BAC = 60^\circ ∠BAC=∠BCA=30∘\angle BAC = \angle BCA = 30^\circ
  5. Use the Sine Rule to find the length of $|AB|$.

    The Sine Rule states asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B}.
    We want to find ∣AB∣|AB|. We know ∣AC∣=12|AC| = 12 cm.

    ∣AB∣sin⁡(∠BCA)=∣AC∣sin⁡(∠ABC)\frac{|AB|}{\sin(\angle BCA)} = \frac{|AC|}{\sin(\angle ABC)} ∣AB∣sin⁡(30∘)=12 cmsin⁡(120∘)\frac{|AB|}{\sin(30^\circ)} = \frac{12 \text{ cm}}{\sin(120^\circ)}
  6. Substitute the sine values and solve for $|AB|$.

    We know sin⁡(30∘)=12\sin(30^\circ) = \frac{1}{2} and sin⁡(120∘)=sin⁡(180∘−60∘)=sin⁡(60∘)=32\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}.

    ∣AB∣12=1232\frac{|AB|}{\frac{1}{2}} = \frac{12}{\frac{\sqrt{3}}{2}} ∣AB∣=12×1232|AB| = \frac{1}{2} \times \frac{12}{\frac{\sqrt{3}}{2}} ∣AB∣=123|AB| = \frac{12}{\sqrt{3}}

    Rationalize the denominator:

    ∣AB∣=1233=43 cm|AB| = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \text{ cm}
  7. Approximate the value.

    ∣AB∣≈4×1.732=6.928 cm|AB| \approx 4 \times 1.732 = 6.928 \text{ cm}

    Rounding to one decimal place, this is 6.9 cm.

    The correct option is A.

    A. 6.9cm\boxed{\text{A. } 6.9\text{cm}}

    : The probability that Saka will score for his team Arsenal is 35\frac{3}{5} and the probability that Rashford will score for his team Manchester United is 13\frac{1}{3}. In a game involving the two teams, find the probability that one of the two players will score.

  8. Define the probabilities of scoring and not scoring for each player.

    Let P(S)P(S) be the probability Saka scores = 35\frac{3}{5}.
    Let P(R)P(R) be the probability Rashford scores = 13\frac{1}{3}.
    The probability Saka does not score, P(S′)=1−P(S)=1−35=25P(S') = 1 - P(S) = 1 - \frac{3}{5} = \frac{2}{5}.
    The probability Rashford does not score, P(R′)=1−P(R)=1−13=23P(R') = 1 - P(R) = 1 - \frac{1}{3} = \frac{2}{3}.

  9. Identify the scenarios where "one of the two players will score".

    This means exactly one player scores. There are two mutually exclusive scenarios:

    • Saka scores AND Rashford does NOT score: P(S and R′)P(S \text{ and } R')
    • Saka does NOT score AND Rashford scores: P(S′ and R)P(S' \text{ and } R)
  10. Calculate the probability for each scenario (assuming the events are independent).

    P(S and R′)=P(S)×P(R′)=35×23=615P(S \text{ and } R') = P(S) \times P(R') = \frac{3}{5} \times \frac{2}{3} = \frac{6}{15} P(S′ and R)=P(S′)×P(R)=25×13=215P(S' \text{ and } R) = P(S') \times P(R) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}
  11. Add the probabilities of these two scenarios to find the total probability.

    P(exactly one scores)=P(S and R′)+P(S′ and R)=615+215=815P(\text{exactly one scores}) = P(S \text{ and } R') + P(S' \text{ and } R) = \frac{6}{15} + \frac{2}{15} = \frac{8}{15}

    The correct option is C.

    C. 815\boxed{\text{C. } \frac{8}{15}}

    : If y=23x3−5x2+3y = \frac{2}{3}x^3 - 5x^2 + 3, find dydx\frac{dy}{dx}

  12. Differentiate each term of the function with respect to xx using the power rule ddx(axn)=anxn−1\frac{d}{dx}(ax^n) = anx^{n-1}.

    For the first term, ddx(23x3)\frac{d}{dx}\left(\frac{2}{3}x^3\right):

    23×3x3−1=2x2\frac{2}{3} \times 3x^{3-1} = 2x^2
  13. Differentiate the second term.

    For the second term, ddx(−5x2)\frac{d}{dx}(-5x^2):

    −5×2x2−1=−10x-5 \times 2x^{2-1} = -10x
  14. Differentiate the third term.

    For the third term, ddx(3)\frac{d}{dx}(3):
    The derivative of a constant is 0.

  15. Combine the derivatives of the terms to get the final derivative.

    dydx=2x2−10x+0\frac{dy}{dx} = 2x^2 - 10x + 0 dydx=2x2−10x\frac{dy}{dx} = 2x^2 - 10x

    The correct option is D.

    D. 2x2−10x\boxed{\text{D. } 2x^2 - 10x}
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