**Introduction to calculus exam preparation problems and answers :**

Calculus is widely used in mathematics, science, and engineering. Calculus deals with limits, differentiation, and integration properties. Calculus is used for comparing the quantities of linear equations. Calculus mainly deals with integration and differentiation processes. It has many problems in differentiation and integration . The quantities of variables are recurrently changed by limits. For calculus exams we need to prepare integration and differentiation problems.

**Ex 1:**

Differentiate the given equation with respect to **t**. **y = 5t ^{4} - 2t^{3} + 8t^{2} + 46t.**

**Sol:**

Given y = 5t^{4} - 2t^{3} + 8t^{2} + 46t.

Differentiating both sides, we have

dy = (5 * 4)t^{(4 - 1)} dt - (2 * 3)t^{(3 - 1)} dt + (8 * 2)t^{(2 - 1)} dt + 46t^{(1 - 1)}
dt.

= 20t^{3} dt - 6t^{2} dt + 16t dt + 46 dt.

= (20t^{3} - 6t^{2} + 16t + 46) dt.

dy / dt = 20t^{3} - 6t^{2} + 16t + 46.

**Answer:**

**d/dt(5t ^{4} - 2t^{3} + 8t^{2} + 46t) = 20t^{3} - 6t^{2} + 16t + 46.**

**Ex 2:**

If **t = x ^{3} - 4x^{2} + 5**. Find

**Sol:**

First step to find dt/dx. So, differentiate the given equation

Given t = x^{3} - 4x^{2} + 5.

dt = 3x^{(3 - 1)} dx - (4 * 2)x^{(2 - 1)} dx + 0 dx.

= 3x^{2} dx - 8x dx.

= (3x^{2} - 8x) dx.

dt / dx = 3x^{2} - 8x.

Then, once again differentiate the value of (dt/dx )with respect to x.

we get (3 * 2)x^{(2 - 1)} - 8x^{(1 - 1)}

= 6x dx - 8

= (6x - 8)

Hence d^{2}t/dx^{2} = 6x - 8.

**Answer:**

**The second differentiation is 6x - 8.**

**Integrate ∫ (3x ^{2} - 4x^{4} + 3x) dx**

**Sol:**

Given ∫ (3x^{2} - 4x^{4} + 3x) dx.

∫ (3x^{2} - 4x^{4} + 3x) dx = ∫ 3x^{2} dx - ∫ 4x^{4} dx + ∫ 3x dx.

= 3∫x^{2} dx - 4∫x^{4} dx + 3∫x dx.

= 3 (x^{3}/3) - 4 (x^{5}/5) + 3 (x^{2}/2) + c.

= x^{3} - (4/5)x^{5} + (3/2)x^{2} + c.

∫ (3x^{2} - 4x^{4} + 3x) dx = x^{3} - (4/5)x^{5} + (3/2)x^{2} + c.

**Answer:**

**∫ (3x ^{2} - 4x^{4} + 3x) dx = x^{3} - (4/5)x^{5} + (3/2)x^{2} + c.**