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An AI solver for integral calculus can help solve various types of integration problems, including standard integration, definite integrals, and integration using different substitutions such as algebraic, trigonometric, and hyperbolic. It can also solve problems involving trigonometric functions like sin, cos, tan, and cot, as well as products and powers of these functions.
The AI can assist with integration using partial fractions, reduction formulae, and techniques like integration by parts. It can also help with double and triple integrals, numerical integration methods like the trapezoidal rule and Simpson's rule, and finding areas under and between curves.
Moreso, the AI can calculate volumes of solids of revolution, centroids of shapes, and solve first-order differential equations using techniques like separation of variables. With the ability to handle a wide range of integral calculus problems, this AI solver can be a valuable tool for students and professionals working in mathematics and related fields.
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1. \int x^2 dx
2. \int 5x dx
3. \int \frac{1}{x} dx
4. \int e^x dx
5. \int \frac{1}{x^2} dx
6. \int \sin(x) dx
7. \int \cos(x) dx
8. \int \sin(2x) dx
9. \int \cos(2x) dx
10. \int \tan(3x) dx
11. \int \cot(2x) dx
12. \int x^3 dx
13. \int \frac{1}{x^3} dx
14. \int e^{2x} dx
15. \int \sin^2(x) dx
16. \int \cos^2(x) dx
17. \int \sin(x) \cos(x) dx
18. \int \sin(3x) dx
19. \int \cos(4x) dx
20. \int \sin(2x) \cos(2x) dx
21. \int \sin(\theta) d\theta
22. \int \cos(\theta) d\theta
23. \int \sin^2(\theta) d\theta
24. \int \cos^2(\theta) d\theta
25. \int \sin(\theta) \cos(\theta) d\theta
26. \int \tan(\theta) d\theta
27. \int \cot(\theta) d\theta
28. \int \sin(3\theta) d\theta
29. \int \cos(4\theta) d\theta
30. \int \sin(2\theta) \cos(2\theta) d\theta
31. \int \sinh(x) dx
32. \int \cosh(x) dx
33. \int \sinh(2x) dx
34. \int \cosh(3x) dx
35. \int \sinh(x) \cosh(x) dx
36. \int \sinh(\theta) d\theta
37. \int \cosh(\theta) d\theta
38. \int \sinh^2(\theta) d\theta
39. \int \cosh^2(\theta) d\theta
40. \int \sinh(\theta) \cosh(\theta) d\theta
41. \int e^{3x} dx
42. \int 2x^2 dx
43. \int \frac{1}{x^4} dx
44. \int e^{-x} dx
45. \int \cos(3x) dx
46. \int \tan^2(x) dx
47. \int \cot^3(x) dx
48. \int x^4 dx
49. \int \sin^3(x) dx
50. \int \cos(5x) dx
The AI Calculus Solver for integral calculus is equipped to handle:
Basic Integration: Polynomial, trigonometric, exponential, and logarithmic functions.
Definite and Indefinite Integrals: Solving for exact values or functions with constants of integration.
Techniques of Integration:
Substitution method.
Integration by parts.
Integration using partial fractions.
Trigonometric substitution.
Applications of Integrals:
Area under a curve.
Volume of solids of revolution (using the disc or shell method).
Accumulated change.
Problem:
Evaluate \( \int \sin^2(x) \, dx \).
Solution
1. Use the trigonometric identity:
\[
\sin^2(x) = \frac{1 – \cos(2x)}{2}.
\]
So,
\[
\int \sin^2(x) \, dx = \int \frac{1 – \cos(2x)}{2} \, dx.
\]
2. Split the integral:
\[
\int \sin^2(x) \, dx = \frac{1}{2} \int 1 \, dx – \frac{1}{2} \int \cos(2x) \, dx.
\]
3. Solve each term:
– \( \frac{1}{2} \int 1 \, dx = \frac{x}{2} \).
– For \( \frac{1}{2} \int \cos(2x) \, dx \), use substitution: Let \( u = 2x \), so \( du = 2 \, dx \).
\[
\int \cos(2x) \, dx = \frac{1}{2} \int \cos(u) \, du = \frac{1}{2} \sin(u) = \frac{1}{2} \sin(2x).
\]
4. Combine results:
\[
\int \sin^2(x) \, dx = \frac{x}{2} – \frac{\sin(2x)}{4} + C.
\]
Answer:
\[
\int \sin^2(x) \, dx = \frac{x}{2} – \frac{\sin(2x)}{4} + C.
\]
Problem
Evaluate \( \int_0^1 e^{2x} \, dx \).
Solution
1. Identify the rule: Use the formula \( \int e^{ax} \, dx = \frac{e^{ax}}{a} + C \).
2. Solve the indefinite integral:
\[
\int e^{2x} \, dx = \frac{e^{2x}}{2}.
\]
3. Apply the limits (from 0 to 1):
\[
\int_0^1 e^{2x} \, dx = \left[ \frac{e^{2x}}{2} \right]_0^1.
\]
4. Evaluate:
\[
\int_0^1 e^{2x} \, dx = \frac{e^2}{2} – \frac{e^0}{2} = \frac{e^2}{2} – \frac{1}{2}.
\]
Answer:
\[
\int_0^1 e^{2x} \, dx = \frac{e^2 – 1}{2}.
\]
Problem
Evaluate \( \int x \sqrt{1 + x^2} \, dx \).
Solution
1. Use substitution: Let \( u = 1 + x^2 \), so \( du = 2x \, dx \).
2. Rewrite the integral:
\[
\int x \sqrt{1 + x^2} \, dx = \frac{1}{2} \int \sqrt{u} \, du.
\]
3. Solve \( \int \sqrt{u} \, du \):
\[
\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}.
\]
4. Back-substitute \( u = 1 + x^2 \):
\[
\frac{1}{2} \cdot \frac{2}{3} (1 + x^2)^{3/2} = \frac{1}{3} (1 + x^2)^{3/2} + C.
\]
Answer:
\[
\int x \sqrt{1 + x^2} \, dx = \frac{1}{3} (1 + x^2)^{3/2} + C.
\]
Problem
Evaluate \( \int x e^x \, dx \).
Solution
1. Use the formula for integration by parts:
\[
\int u \, dv = uv – \int v \, du.
\]
Let \( u = x \) and \( dv = e^x dx \).
2. Find \( du \) and \( v \):
– \( du = dx \),
– \( v = \int e^x dx = e^x \).
3. Substitute into the formula:
\[
\int x e^x \, dx = x e^x – \int e^x \, dx.
\]
4. Solve \( \int e^x \, dx \):
\[
\int x e^x \, dx = x e^x – e^x + C.
\]
5. Factorize:
\[
\int x e^x \, dx = e^x (x – 1) + C.
\]
Answer
\[
\int x e^x \, dx = e^x (x – 1) + C.
\]
Problem
Find the area under \( y = x^2 \) between \( x = 0 \) and \( x = 3 \).
Solution
1. Set up the integral:
\[
\int_0^3 x^2 \, dx.
\]
2. Solve the indefinite integral:
\[
\int x^2 \, dx = \frac{x^3}{3}.
\]
3. Apply the limits:
\[
\int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3.
\]
4. Evaluate:
\[
\int_0^3 x^2 \, dx = \frac{3^3}{3} – \frac{0^3}{3} = \frac{27}{3} – 0 = 9.
\]
Answer
The area under the curve is \( 9 \) square units.
Problem
Find the volume of the solid obtained by rotating \( y = x^2 \) about the x-axis between \( x = 0 \) and \( x = 2 \).
Solution
1. Use the formula for volume of revolution:
\[
V = \pi \int_a^b [f(x)]^2 \, dx.
\]
2. Substitute \( f(x) = x^2 \):
\[
V = \pi \int_0^2 (x^2)^2 \, dx = \pi \int_0^2 x^4 \, dx.
\]
3. Solve the indefinite integral:
\[
\int x^4 \, dx = \frac{x^5}{5}.
\]
4. Apply the limits:
\[
V = \pi \left[ \frac{x^5}{5} \right]_0^2 = \pi \left( \frac{2^5}{5} – \frac{0^5}{5} \right).
\]
5. Simplify:
\[
V = \pi \left( \frac{32}{5} –
0 \right) = \frac{32\pi}{5}.
\]
Answer:
The volume of the solid is \( \frac{32\pi}{5} \).
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