- \(\displaystyle \int \sqrt {4-x} ~dx = \color{blue} {- \frac {2}{3}(4-x)^{\frac {3}{2}}}\)
Solution: \(u = 4 - x\), \(du = - ~dx\)
Solution: \(\sqrt {4-x} ~dx = - u^{\frac {1}{2}} ~du\)
Solution
Take \(x \le 4\).
1. Make the substitution \(u = 4 - x\).Then \(du = -dx \implies dx = -du\).
The substitution rule for indefinite integrals states that if \(u = g(x)\) is differentiable and \(F\) is an antiderivative of \(f\), then \(\int f(g(x))\, g'(x)\, dx = F(g(x)) + C\). Here \(g(x) = 4 - x\) with \(g'(x) = -1\), and \(f(u) = u^{1/2}\), so the substitution is valid.
2. Transform and evaluate the integral: \[\int \sqrt{4-x} \, dx = \int u^{1/2} (-du) = -\int u^{1/2} \, du = -\frac{2}{3} u^{3/2} + C.\]
2.1. The substitution \(u = 4 - x\) from 1 gives \(dx = -du\), so \(\sqrt{4-x}\,dx = u^{1/2}(-du)\).
2.2. The constant factor \(-1\) pulls out of the integral: \(\int u^{1/2}(-du) = -\int u^{1/2}\,du\).
2.3. The power rule \(\int u^p\,du = \frac{u^{p+1}}{p+1} + C\) applies with \(p = 1/2\), giving \(-\frac{u^{3/2}}{3/2} + C = -\frac{2}{3}u^{3/2} + C\).
3. Substitute back \(u = 4-x\): \[\int \sqrt{4-x} \, dx = -\frac{2}{3} (4-x)^{3/2} + C.\]
Replacing \(u\) by \(4 - x\) in the antiderivative \(-\frac{2}{3}u^{3/2} + C\) from 2 yields \(-\frac{2}{3}(4-x)^{3/2} + C\).