13

  1. (a)

    Find ∫(32+7x+sin⁡5x) dx.\displaystyle\int\left(\frac{3}{\sqrt{2+7x}}+\sin 5x\right)\,dx.

  2. (b)

    If f′(x)=2xf'(x)=2x and g′(x)=2xg'(x)=2x, can we conclude that f(x)=g(x)f(x)=g(x)? Give a reason for your answer.

  3. (c)

    Given that V=20−3exV=20-3e^x, explain why the value of VV can never be 20.

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Worked solution

13(a)

1. Rewrite the integrand using fractional indices to prepare for integration

∫(3(2+7x)−12+sin⁡5x) dx\int \left( 3(2+7x)^{-\frac{1}{2}} + \sin 5x \right) \, dx

2. Integrate the power term with a linear argument

∫3(2+7x)−12 dx=3(2+7x)1212×7=672+7x\int 3(2+7x)^{-\frac{1}{2}} \, dx = \frac{3(2+7x)^{\frac{1}{2}}}{\frac{1}{2} \times 7} = \frac{6}{7}\sqrt{2+7x}

3. Integrate the trigonometric term and combine with the constant of integration

∫sin⁡5x dx=−15cos⁡5x\int \sin 5x \, dx = -\frac{1}{5}\cos 5x

Combining the results:
=672+7x−15cos⁡5x+C= \frac{6}{7}\sqrt{2+7x} - \frac{1}{5}\cos 5x + C

Answer

672+7x−15cos⁡5x+C\frac{6}{7}\sqrt{2+7x} - \frac{1}{5}\cos 5x + C

13(b)

1. Integrate the given derivatives to find the general forms of the functions

Since f′(x)=2xf'(x) = 2x, integrating gives f(x)=x2+C1f(x) = x^2 + C_1.
Similarly, since g′(x)=2xg'(x) = 2x, integrating gives g(x)=x2+C2g(x) = x^2 + C_2.

2. Explain why the functions are not necessarily equal

The constants of integration C1C_1 and C2C_2 are arbitrary and can be different. Therefore, f(x)f(x) and g(x)g(x) are not necessarily the same function; they can differ by a constant.

Answer

No, we cannot conclude that f(x)=g(x)f(x)=g(x) because they can differ by a constant of integration.

13(c)

1. State the property of the exponential function

For all real values of xx, ex>0e^x > 0.

2. Deduce the restriction on V

Since ex>0e^x > 0, it follows that 3ex>03e^x > 0.
Therefore, V=20−3ex<20V = 20 - 3e^x < 20.
Hence, VV can never be equal to 20.

Answer

Since ex>0e^x > 0 for all real xx, 3ex>03e^x > 0. Therefore, V=20−3ex<20V = 20 - 3e^x < 20, which means VV can never be exactly 20.

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