Show that ddx[(2x)(3−5x)65]=(6−22x)(3−5x)15\frac{d}{dx} \left[ (2x)(3-5x)^{\frac{6}{5}} \right] = (6-22x)(3-5x)^{\frac{1}{5}}. Hence, evaluate ∫x(3−5x)15dx\int x(3-5x)^{\frac{1}{5}} dx.

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Worked solution

11(a)

1. Apply the product rule and chain rule to differentiate the expression.

Let y=(2x)(3−5x)65y = (2x)(3-5x)^{\frac{6}{5}}.
Using the product rule ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}:
dydx=(2x)⋅ddx[(3−5x)65]+(3−5x)65⋅ddx(2x)\frac{dy}{dx} = (2x) \cdot \frac{d}{dx}\left[(3-5x)^{\frac{6}{5}}\right] + (3-5x)^{\frac{6}{5}} \cdot \frac{d}{dx}(2x)
dydx=(2x)[65(3−5x)15(−5)]+(3−5x)65(2)\frac{dy}{dx} = (2x) \left[ \frac{6}{5}(3-5x)^{\frac{1}{5}}(-5) \right] + (3-5x)^{\frac{6}{5}}(2)

2. Simplify the expression by factoring out the common term.

dydx=−12x(3−5x)15+2(3−5x)65\frac{dy}{dx} = -12x(3-5x)^{\frac{1}{5}} + 2(3-5x)^{\frac{6}{5}}
Factor out (3−5x)15(3-5x)^{\frac{1}{5}}:
dydx=(3−5x)15[−12x+2(3−5x)1]\frac{dy}{dx} = (3-5x)^{\frac{1}{5}} \left[ -12x + 2(3-5x)^{1} \right]

3. Expand and simplify the terms inside the bracket to obtain the final result.

dydx=(3−5x)15[−12x+6−10x]\frac{dy}{dx} = (3-5x)^{\frac{1}{5}} \left[ -12x + 6 - 10x \right]
dydx=(3−5x)15(6−22x)\frac{dy}{dx} = (3-5x)^{\frac{1}{5}} (6 - 22x)
dydx=(6−22x)(3−5x)15\frac{dy}{dx} = (6-22x)(3-5x)^{\frac{1}{5}}

Answer

ddx[(2x)(3−5x)65]=(6−22x)(3−5x)15\frac{d}{dx} \left[ (2x)(3-5x)^{\frac{6}{5}} \right] = (6-22x)(3-5x)^{\frac{1}{5}} (Shown)

11(b)

1. Express the integral of the result from part (a) as the original function.

From part (a), we have:
∫(6−22x)(3−5x)15 dx=2x(3−5x)65+C\int (6-22x)(3-5x)^{\frac{1}{5}} \, dx = 2x(3-5x)^{\frac{6}{5}} + C

2. Rewrite the integrand to isolate the required integral.

Notice that 6−22x=2(3−5x)−12x6-22x = 2(3-5x) - 12x. Therefore:
(6−22x)(3−5x)15=[2(3−5x)−12x](3−5x)15(6-22x)(3-5x)^{\frac{1}{5}} = [2(3-5x) - 12x](3-5x)^{\frac{1}{5}}
=2(3−5x)65−12x(3−5x)15= 2(3-5x)^{\frac{6}{5}} - 12x(3-5x)^{\frac{1}{5}}

Substituting this into the integral equation:
∫[2(3−5x)65−12x(3−5x)15] dx=2x(3−5x)65+C\int \left[ 2(3-5x)^{\frac{6}{5}} - 12x(3-5x)^{\frac{1}{5}} \right] \, dx = 2x(3-5x)^{\frac{6}{5}} + C
∫2(3−5x)65 dx−12∫x(3−5x)15 dx=2x(3−5x)65+C\int 2(3-5x)^{\frac{6}{5}} \, dx - 12 \int x(3-5x)^{\frac{1}{5}} \, dx = 2x(3-5x)^{\frac{6}{5}} + C

3. Evaluate the standard integral and solve for the required integral.

Evaluate ∫2(3−5x)65 dx\int 2(3-5x)^{\frac{6}{5}} \, dx:
2[(3−5x)115115×(−5)]=−211(3−5x)1152 \left[ \frac{(3-5x)^{\frac{11}{5}}}{\frac{11}{5} \times (-5)} \right] = -\frac{2}{11}(3-5x)^{\frac{11}{5}}

Substitute back:
−211(3−5x)115−12∫x(3−5x)15 dx=2x(3−5x)65+C-\frac{2}{11}(3-5x)^{\frac{11}{5}} - 12 \int x(3-5x)^{\frac{1}{5}} \, dx = 2x(3-5x)^{\frac{6}{5}} + C
12∫x(3−5x)15 dx=−211(3−5x)115−2x(3−5x)65+C′12 \int x(3-5x)^{\frac{1}{5}} \, dx = -\frac{2}{11}(3-5x)^{\frac{11}{5}} - 2x(3-5x)^{\frac{6}{5}} + C'
∫x(3−5x)15 dx=−166(3−5x)115−16x(3−5x)65+C′′\int x(3-5x)^{\frac{1}{5}} \, dx = -\frac{1}{66}(3-5x)^{\frac{11}{5}} - \frac{1}{6}x(3-5x)^{\frac{6}{5}} + C''

4. Factorise and simplify the final answer.

∫x(3−5x)15 dx=(3−5x)65[−166(3−5x)−16x]+C\int x(3-5x)^{\frac{1}{5}} \, dx = (3-5x)^{\frac{6}{5}} \left[ -\frac{1}{66}(3-5x) - \frac{1}{6}x \right] + C
=(3−5x)65[−122+566x−1166x]+C= (3-5x)^{\frac{6}{5}} \left[ -\frac{1}{22} + \frac{5}{66}x - \frac{11}{66}x \right] + C
=(3−5x)65[−122−666x]+C= (3-5x)^{\frac{6}{5}} \left[ -\frac{1}{22} - \frac{6}{66}x \right] + C
=(3−5x)65[−122−111x]+C= (3-5x)^{\frac{6}{5}} \left[ -\frac{1}{22} - \frac{1}{11}x \right] + C
=−122(1+2x)(3−5x)65+C= -\frac{1}{22}(1+2x)(3-5x)^{\frac{6}{5}} + C

Answer

−122(1+2x)(3−5x)65+C-\frac{1}{22}(1+2x)(3-5x)^{\frac{6}{5}} + C

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