Integral calculus is a branch of mathematics used to find the total or accumulated value of quantities, such as area, distance, or volume. It is mainly based on the process of integration.
Example: Integration can be used to find the area under a curve or the distance travelled from a given speed.
Solved Examples
Example 1: Solve: ∫x9dx
Answer:
∫x9dx = [x9 + 1 / (9 + 1)] + C
⇒ ∫x9dx = (x10 / 10) + C
Example 2: Solve ∫ 2x + 3dx.
Answer:
∫ 2x + 3dx = ∫ 2x 23dx
⇒ ∫ 2x + 3dx = 8∫ 2x dx
⇒ ∫ 2x + 3dx = 8[2x / loge2] + C
Example 3: Evaluate: ∫(x3 + 3x2 + 5x + 6)dx
Answer:
∫(x3 + 3x2 + 5x + 6) dx = ∫x3dx + ∫ 3x2dx + ∫ 5x dx + ∫ 6dx
⇒ ∫(x3 + 3x2 + 5x + 6) dx = (x3+1 / (3+1)) + 3[(x2+1 / (2+1))] + 5[(x1+1 / (1+1))] + 6[x0+1/ (0+1)] + C
⇒ ∫(x3 + 3x2 + 5x + 6) dx = (x4 / 4) + x3 + (5/2)x2 + 6x + C
Example 4: Solve ∫(x + 6cos x) dx.
Answer:
∫(x + 6cos x) dx = ∫x dx + ∫cos x dx
⇒ ∫(x + 6cos x) dx = ∫x dx + 6 ∫cos x dx
⇒ ∫(x + 6cos x) dx = [x1+1 / (1+1)] + 6sin x + C
⇒ ∫(x + 6cos x) dx = [x2 / 2] + 6sin x + C
Example 5: Evaluate ∫[1/{4√(x2 - 16)}
Answer:
∫[1/{4√(x2 - 16)}] = ∫[1/{4√(x2 - 42)}]
⇒ ∫ [1 / {x√(x2 - a2)}] dx = (1/a) sec-1(x/a) + C
⇒ ∫[1/{4√(x2 - 16)}] = (1/4) sec-1(x/4) + C
Practice Problems
1. Calculate the integral of the function: ∫ (3x2 - 5x + 4) dx
2. Evaluate the integral involving a basic trigonometric function: ∫cos(2x) dx
3. Use substitution to solve the integral: ∫x sin(x2) dx
4. Calculate the integral using the method of integration by parts: ∫x2 eˣ dx
5. Find the area under the curve from x = 0 to x = ?. for the function: ∫ 0pi Sin(x) dx
