6. Given that \( \frac{d y}{d x}=2 x+7 \) and \( y=5 \) when \( x=-1 \), find the value of \( y \) when \( x=2 \) 7. Given that \( \frac{d v}{d t}=3-2 t \) and \( v=2 \) when \( t=1 \), find the value of \( v \) when \( t=2 \)

To solve the first problem, we need to integrate \(\frac{d y}{d x} = 2x + 7\). The integral is: \[ y = x^2 + 7x + C \] To find \(C\), we use the condition \(y=5\) when \(x=-1\): \[ 5 = (-1)^2 + 7(-1) + C \implies 5 = 1 - 7 + C \implies C = 11 \] Thus, the equation for \(y\) is: \[ y = x^2 + 7x + 11 \] Now, substituting \(x=2\): \[ y = 2^2 + 7(2) + 11 = 4 + 14 + 11 = 29 \] For the second problem, we need to integrate \(\frac{d v}{d t} = 3 - 2t\): \[ v = 3t - t^2 + C \] Using the condition \(v=2\) when \(t=1\): \[ 2 = 3(1) - (1)^2 + C \implies 2 = 3 - 1 + C \implies C = 0 \] Thus, we have \(v = 3t - t^2\). Now substituting \(t=2\): \[ v = 3(2) - (2)^2 = 6 - 4 = 2 \] So, when \(x=2\), \(y = 29\) and when \(t=2\), \(v = 2\).