Use integration by substitution to find

∫
0
1
1
1
+
𝑎
𝑥
2
𝑑
𝑥
where a=19

Enter your answer to two decimal places, rounding appropriately.

Question

Use integration by substitution to find

∫
0
1
1
1
+
𝑎
𝑥
2
𝑑
𝑥
 where a=19

Enter your answer to two decimal places, rounding appropriately.

BlackTom AI · Accepted Answer

Let t = √a x so that dx = dt/√a and with a = 19 we have t = √19 x. The integral becomes
∫_0^1 1/(1 + 19 x^2) dx = ∫_{t=0}^{t=√19} (1/√19) · 1/(1 + t^2) dt.
Relevant fact: d/dt [arctan t] = 1/(1 + t^2), so the antiderivative of 1/(1 + t^2) is arctan t. Applying that,
(1/√19) ∫_0^{√19} 1/(1 + t^2) dt = (1/√19) [arctan t]_0^{√19} = (1/√19) · arctan(√19).

Compute the numerical value step by step:
√19 ≈ 4.358898943.
1/√19 ≈ 0.2294157339.
To evaluate arctan(√19) accurately use the identity arctan(√19) = π/2 − arctan(1/√19). With y = 1/√19 ≈ 0.2294157339, a series expansion gives arctan(y) ≈ y − y^3/3 + y^5/5 − y^7/7 ≈ 0.2255144. Thus arctan(√19) ≈ π/2 − 0.2255144 ≈ 1.3452819.

Multiply by 1/√19:
(1/√19) · arctan(√19) ≈ 0.2294157339 × 1.3452819 ≈ 0.3086288.

Round to two decimal places (rounding appropriately): 0.31.

Why this is correct: the substitution reduced the integrand to the standard arctangent integrand 1/(1 + t^2), whose antiderivative is arctan t. The factor 1/√a accounts for the change of variable. There are no other choices offered here; any alternative numeric value that does not equal (1/√19)·arctan(√19) (≈ 0.3086) would be incorrect.

Final answer (rounded to two decimals): 0.31, which matches the provided answer.

Use integration by substitution to find ∫ 0 1 1 1 + 𝑎 𝑥 2 𝑑 𝑥 where a=19 Enter your answer to two decimal places, rounding appropriately.简答题

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