(a) Solve : (2^{3y + 2} – 7(2^{2y + 2}) – 31(2^{y}) – 8 = 0, y in R).
(b) Find (int (sqrt{x^{2} + 1}) xdx).
Explanation
(a) (2^{3y + 2} – 7(2^{2y + 2}) – 31(2^{y}) – 8 = 0)
(equiv (2^{2})(2^{y})^{3} – 7(2^{2})(2^{y})^{2} – 31(2^{y}) – 8 = 0)
Let (2^{y} = x),
(4x^{3} – 28x^{2} – 31x – 8 = 0)
If x = 8, (4(8^{3}) – 28(8^{2}) – 31(8) – 8 = 2048 – 1792 – 248 – 8)
= (0)
(therefore (x – 8) text{is a factor})
Dividing (frac{4x^{3} – 28x^{2} – 31x – 8}{x – 8}), we get
= (4x^{2} + 4x + 1 )
= ((2x + 1)^{2})
Therefore, (x = -frac{1}{2}, -frac{1}{2}, 8)
Recall, (x = 2^{y})
(2^{y} = -frac{1}{2}) has no real solution.
(2^{y} = 8 = 2^{3} implies y = 3)
(b) (int (sqrt{x^{2} + 1}) xdx)
Let (u^{2} = x^{2} + 1; 2frac{mathrm d u}{mathrm d x} = 2x)
(mathrm d u = x mathrm d x)
(therefore int (sqrt{x^{2} + 1}) xdx = int (sqrt{u^{2}}) mathrm d u)
= (int u mathrm d u)
= (frac{u^{2}}{2} + c)
= (frac{x^{2} + 1}{2} + c)