Question

Example. Compute: $$int_{0}^{16} sqrt{4-sqrt{x}} dx$$ Explanation. While it is not obvious at all, let us try the substitution $$u = sqrt{x}.$$ Then $$du = oxed{?} dx,$$ $$dx = 2sqrt{x} du = 2u du,$$ and so $$int_{0}^{16} sqrt{4-sqrt{x}} dx = int_{g(0)}^{g(16)} oxed{?} du$$ $$= int_{oxed{?}}^{oxed{?}} oxed{?} du.$$ From here we now make the second (and more obvious) substitution $$w = 4 - u.$$ Then $u = 4 - w$, and $$dw = -du,$$ $$du = -dw.$$ So $$int_{0}^{16} sqrt{4-sqrt{x}} dx = int_{0}^{4} 2usqrt{4 - u} du$$ $$= int_{w(0)}^{w(4)} 2(4 - w)sqrt{w}(-1) dw$$ $$= -int_{oxed{?}}^{oxed{?}} (8w^{frac{1}{2}} - 2w^{frac{3}{2}}) dw$$ $$= int_{oxed{?}}^{oxed{?}} (8w^{frac{1}{2}} - 2w^{frac{3}{2}}) dw$$ $$= left[ oxed{?} ight]_{0}^{4}$$ $$= left( frac{16}{3}(4)^{frac{3}{2}} - frac{4}{5}(4)^{frac{5}{2}} ight) - (0 - 0)$$ $$= frac{128}{3} - frac{128}{5}$$ $$= frac{256}{15}.$$

          Example. Compute:

$$int_{0}^{16} sqrt{4-sqrt{x}} dx$$

Explanation. While it is not obvious at all, let us try the substitution

$$u = sqrt{x}.$$

Then

$$du = oxed{?} dx,$$
$$dx = 2sqrt{x} du = 2u du,$$

and so

$$int_{0}^{16} sqrt{4-sqrt{x}} dx = int_{g(0)}^{g(16)} oxed{?} du$$
$$= int_{oxed{?}}^{oxed{?}} oxed{?} du.$$

From here we now make the second (and more obvious) substitution

$$w = 4 - u.$$

Then $u = 4 - w$, and

$$dw = -du,$$
$$du = -dw.$$

So

$$int_{0}^{16} sqrt{4-sqrt{x}} dx = int_{0}^{4} 2usqrt{4 - u} du$$
$$= int_{w(0)}^{w(4)} 2(4 - w)sqrt{w}(-1) dw$$
$$= -int_{oxed{?}}^{oxed{?}} (8w^{frac{1}{2}} - 2w^{frac{3}{2}}) dw$$
$$= int_{oxed{?}}^{oxed{?}} (8w^{frac{1}{2}} - 2w^{frac{3}{2}}) dw$$
$$= left[ oxed{?} 
ight]_{0}^{4}$$
$$= left( frac{16}{3}(4)^{frac{3}{2}} - frac{4}{5}(4)^{frac{5}{2}} 
ight) - (0 - 0)$$
$$= frac{128}{3} - frac{128}{5}$$
$$= frac{256}{15}.$$

$Example. Compute: int0^16 sqrt4-sqrtx dx Explanation. While it is not obvious at all, let us try the substitution u = sqrtx. Then du = oxed? dx, dx = 2sqrtx du = 2u du, and so int0^16 sqrt4-sqrtx dx = intg(0)^g(16) oxed? du = intoxed?^oxed? oxed? du. From here we now make the second (and more obvious) substitution w = 4 - u. Then u = 4 - w, and dw = -du, du = -dw. So int0^16 sqrt4-sqrtx dx = int0^4 2usqrt4 - u du = intw(0)^w(4) 2(4 - w)sqrtw(-1) dw = -intoxed?^oxed? (8w^frac12 - 2w^frac32) dw = intoxed?^oxed? (8w^frac12 - 2w^frac32) dw = left[ oxed? ight]0^4 = left( frac163(4)^frac32 - frac45(4)^frac52 ight) - (0 - 0) = frac1283 - frac1285 = frac25615.$

Added by Bill K.

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