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Determine the following integrals: 1. integral(1)/(-2x2+20x-47)dx 2. integral3x2xdx 3. integral e⁻²x[3]x6dx 4. integral5 4x dx 5. integral⁻¹((7x)/(3))dx

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  1. Find the first derivatives of xx and yy with respect to tt.

    Given x=t2x = t^2 and y=2t6y = 2t^6.

    dxdt=ddt(t2)=2t\frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t dydt=ddt(2t6)=12t5\frac{dy}{dt} = \frac{d}{dt}(2t^6) = 12t^5

  2. Calculate the first derivative dydx\frac{dy}{dx} using the chain rule.

    dydx=dy/dtdx/dt=12t52t=6t4\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{12t^5}{2t} = 6t^4

  3. Calculate the second derivative d2ydx2\frac{d^2y}{dx^2} using the chain rule.

    d2ydx2=ddx(dydx)=ddt(dydx)⋅dtdx\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}

    First, find ddt(dydx)\frac{d}{dt}\left(\frac{dy}{dx}\right):

    ddt(6t4)=24t3\frac{d}{dt}(6t^4) = 24t^3

    Next, find dtdx\frac{dt}{dx}:

    dtdx=1dx/dt=12t\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2t}

    Now, multiply these two expressions:

    d2ydx2=(24t3)⋅(12t)=12t2\frac{d^2y}{dx^2} = (24t^3) \cdot \left(\frac{1}{2t}\right) = 12t^2

  4. Substitute the given value of t=13t = \frac{1}{3} into the expression for d2ydx2\frac{d^2y}{dx^2}.

    d2ydx2∣t=13=12(13)2=12(19)=129=43\frac{d^2y}{dx^2} \Big|_{t=\frac{1}{3}} = 12\left(\frac{1}{3}\right)^2 = 12\left(\frac{1}{9}\right) = \frac{12}{9} = \frac{4}{3}

    The value of d2ydx2\frac{d^2y}{dx^2} at t=13t = \frac{1}{3} is 43\boxed{\frac{4}{3}}.

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