now we integrate term by term.\n∫e^(-4x²) dx = ∫(1 - 4x²/1! + 16x⁴/2! - 64x⁶/3! +... + (-1)ⁿ4ⁿx²ⁿ/n!)dx\n= c…

now we integrate term by term.\n∫e^(-4x²) dx = ∫(1 - 4x²/1! + 16x⁴/2! - 64x⁶/3! +... + (-1)ⁿ4ⁿx²ⁿ/n!)dx\n= c + x - 4x³/(3·1!) + 16x⁵/(5·2!) - 64x⁷/(7·3!) +... + (-1)ⁿ4ⁿx²ⁿ⁺¹/((2n + 1)·n!) +...\nthis series converges for all x because the original series for e^(-4x²) converges for all x.\n(b) the fundamental theorem of calculus gives\n∫₀⁰.⁵e^(-4x²) dx = x - 4x³/3 + 16x⁵/(5·2!) - 64x⁷/(7·3!) +...₀⁰.⁵\n= 0.5 - 1/6 + 1/20 - 1/84 + 1/432 -...\n≈ - 1/6 + - 1/84 + 1/432 ≈ \nthe alternating series estimation theorem shows that the error involved in this approximation is less than 0.001.\nneed help?

now we integrate term by term.\n∫e^(-4x²) dx = ∫(1 - 4x²/1! + 16x⁴/2! - 64x⁶/3! +... + (-1)ⁿ4ⁿx²ⁿ/n!)dx\n= c + x - 4x³/(3·1!) + 16x⁵/(5·2!) - 64x⁷/(7·3!) +... + (-1)ⁿ4ⁿx²ⁿ⁺¹/((2n + 1)·n!) +...\nthis series converges for all x because the original series for e^(-4x²) converges for all x.\n(b) the fundamental theorem of calculus gives\n∫₀⁰.⁵e^(-4x²) dx = x - 4x³/3 + 16x⁵/(5·2!) - 64x⁷/(7·3!) +...₀⁰.⁵\n= 0.5 - 1/6 + 1/20 - 1/84 + 1/432 -...\n≈ - 1/6 + - 1/84 + 1/432 ≈ \nthe alternating series estimation theorem shows that the error involved in this approximation is less than 0.001.\nneed help?

Answer

Explanation:

Step1: Recall power - series of $e^u$

The power - series of $e^u=\sum_{n = 0}^{\infty}\frac{u^n}{n!}=1 + u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots$. Let $u=-4x^{2}$, then $e^{-4x^{2}}=\sum_{n = 0}^{\infty}\frac{(-4x^{2})^{n}}{n!}=1-\frac{4x^{2}}{1!}+\frac{16x^{4}}{2!}-\frac{64x^{6}}{3!}+\cdots+(-1)^{n}\frac{4^{n}x^{2n}}{n!}$.

Step2: Integrate term - by - term

$\int e^{-4x^{2}}dx=\int\left(1-\frac{4x^{2}}{1!}+\frac{16x^{4}}{2!}-\frac{64x^{6}}{3!}+\cdots+(-1)^{n}\frac{4^{n}x^{2n}}{n!}\right)dx$. Using the power - rule of integration $\int x^{m}dx=\frac{x^{m + 1}}{m+1}+C$ ($m\neq - 1$), we get $\int e^{-4x^{2}}dx=C + x-\frac{4x^{3}}{3\cdot1!}+\frac{16x^{5}}{5\cdot2!}-\frac{64x^{7}}{7\cdot3!}+\cdots+(-1)^{n}\frac{4^{n}x^{2n + 1}}{(2n+1)\cdot n!}+\cdots$.

Step3: Evaluate the definite integral

For $\int_{0}^{0.5}e^{-4x^{2}}dx=\left[x-\frac{4x^{3}}{3}+\frac{16x^{5}}{5\cdot2!}-\frac{64x^{7}}{7\cdot3!}+\cdots\right]_{0}^{0.5}$. Substitute $x = 0.5$ and $x = 0$: When $x = 0.5$, we have $0.5-\frac{4\times(0.5)^{3}}{3}+\frac{16\times(0.5)^{5}}{5\times2!}-\frac{64\times(0.5)^{7}}{7\times3!}+\cdots$. $0.5-\frac{4\times\frac{1}{8}}{3}+\frac{16\times\frac{1}{32}}{10}-\frac{64\times\frac{1}{128}}{42}+\cdots=0.5-\frac{1}{6}+\frac{1}{20}-\frac{1}{84}+\cdots$.

Answer:

The value of $\int_{0}^{0.5}e^{-4x^{2}}dx\approx0.5-\frac{1}{6}+\frac{1}{20}-\frac{1}{84}+\frac{1}{432}\approx0.4283$ (calculating the sum of the first few terms of the alternating series).