∫_(-2)⁰ {x³ + 3x² + 3x + 3 + (x+1) cos(x+1)} dx is equal to (a) -4 (b) 0 (c) 4 (d) 6

Q: Wait, why do we change the limits of integration? Can't we just integrate with respect to $x$?

You absolutely can integrate with respect to $x$, but it makes the algebra significantly more tedious. By substituting $t = x+1$, we shift the entire coordinate system. This makes the limits symmetric (from $-1$ to $1$), which allows us to use the powerful 'odd and even function' property. It is a classic JEE trick to save time and reduce calculation errors!

Q: How do I know if a function is odd or even quickly?

It is simple! Replace $t$ with $-t$ in your function $f(t)$. If $f(-t) = f(t)$, the function is even (like $\cos t$ or $t^2$). If $f(-t) = -f(t)$, the function is odd (like $t^3$ or $t \cos t$). In this problem, $t^3$ and $t \cos t$ are both odd because $(-t)^3 = -t^3$ and $(-t)\cos(-t) = -t \cos t$. Their sum is also odd, and the integral of any odd function over a symmetric interval $[-a, a]$ is always zero.

Q: Wait, why did the $t^3$ and $t \cos t$ terms just disappear? Is that legal?

It is perfectly legal! Geometrically, an odd function is symmetric about the origin but inverted. This means the area under the curve on the negative side is the exact negative of the area on the positive side. When you sum them up from $-1$ to $1$, they cancel each other out perfectly. It is not that they 'disappear'—they sum to zero!

Question

$\int_{-2}^{0} \{x^3 + 3x^2 + 3x + 3 + (x+1) \cos(x+1)\} dx$ is equal to

Options:

(A) -4

(B) 0

(C) 4

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear students, today we are going to tackle a problem that, at first glance, might make your heart skip a beat. We are looking at the integral:

$$I = \int_{-2}^{0} \{x^3 + 3x^2 + 3x + 3 + (x+1) \cos(x+1)\} dx$$

It looks like a chaotic mess of polynomials and trigonometry, doesn't it? But in the world of JEE Advanced, chaos is often just order in disguise. Let us peel back the layers together.

### The Pattern Hunt

Whenever you see a polynomial like $x^3 + 3x^2 + 3x + 3$, your mathematical intuition should immediately start searching for a pattern. It is screaming to be a binomial expansion!

We know that $(x+1)^3 = x^3 + 3x^2 + 3x + 1$. Our expression has a $+3$ at the end, which is just $1 + 2$. So, we can rewrite the polynomial part as $(x+1)^3 + 2$.

Suddenly, the entire integrand becomes:

$$I = \int_{-2}^{0} \{(x+1)^3 + 2 + (x+1) \cos(x+1)\} dx$$

Do you see the beauty of it? Every variable term now contains $(x+1)$. This is the designer's fingerprint—a clear invitation to simplify.

### The Transformation

Let us make our lives easier. Let $t = x+1$. When we differentiate, we get $dt = dx$.

We must be careful—the limits of integration are the soul of a definite integral. When $x = -2$, $t = -2 + 1 = -1$. When $x = 0$, $t = 0 + 1 = 1$.

Our integral is now transformed into:

$$I = \int_{-1}^{1} \{t^3 + 2 + t \cos t\} dt$$

Look at those limits: $-1$ to $1$. This is the symmetric interval we were hoping for!

### The Symmetry Magic

This is where the JEE magic happens. We have the property:

$$\int_{-a}^{a} f(t) dt = \int_{-a}^{a} f_{even}(t) dt + \int_{-a}^{a} f_{odd}(t) dt$$

We know that the integral of an odd function over a symmetric interval is exactly zero. Let us test our terms.

$f_1(t) = t^3$ is odd because $f_1(-t) = (-t)^3 = -t^3$. $f_2(t) = t \cos t$ is also odd because $f_2(-t) = (-t) \cos(-t) = -t \cos t$. Since both are odd, their integral over $[-1, 1]$ vanishes into thin air!

### The Final Victory

We are left with the simplest integral imaginable:

$$I = \int_{-1}^{1} 2 dt$$

This is just the area of a rectangle with height $2$ and width $2$. Calculating this, we get:

$$[2t]_{-1}^{1} = 2(1) - 2(-1) = 2 + 2 = 4$$

And there you have it! The intimidating monster has been tamed by the elegance of symmetry. Always remember: when the math looks scary, look for the symmetry. It is usually the key to the kingdom. The final answer is **4**.

Animated Solution for Mathematics - Definite Integration: $\int_{- 2}^{0} {x^{3} + 3 x^{2} + 3 x + 3 + (x + 1) cos (x + 1)} d x$ is equal to

Select Answer:

Visualized Solution

The Given Integral

Spotting the Pattern

Algebraic Manipulation

Rewriting the Integral

Applying Substitution

Finding $d t$

Updating Integration Limits

The Simplified Integral

Definite Integral Property

Is $t^{3}$ Odd or Even?

Is $t cos t$ Odd or Even?

Area Cancellation

Integrating the Constant

Final Calculation

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Pattern Hunt

The Transformation

The Symmetry Magic

The Final Victory

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫−20​{x3+3x2+3x+3+(x+1)cos(x+1)}dx is equal to