Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

Bounding an integral over a specific interval can often be accomplished by evaluating the integrand at the function's endpoints to establish an interval of containment. For the integral $I=\int_{1}^{2}\frac{dx}{\sqrt{2x^{3-}9x^{2+}12x+4}}$, we focus on the function $f(x) = (2x^3 - 9x^2 + 12x + 4)^{-1/2}$. By substituting the limits of integration into this expression, we determine the values at the boundaries: $f(1) = (2 - 9 + 12 + 4)^{-1/2} = \frac{1}{3}$ and $f(2) = (16 - 36 + 24 + 4)^{-1/2} = \frac{1}{\sqrt{8}}$.

Since the interval of integration has a width of $1$, the value of the integral $I$ is constrained by the range of the function values at these endpoints. Expressing the integral within these bounds leads to the inequality $\frac{1}{3} < I < \frac{1}{\sqrt{8}}$. Squaring all parts of this inequality to determine the range for $I^2$ results in $\frac{1}{9} < I^2 < \frac{1}{8}$, which defines the interval for the squared value of the integral.

To determine the behavior of a function at a critical point, one must utilize the second derivative test, which requires evaluating the curvature of the function where its first derivative is zero. Applying the Fundamental Theorem of Calculus to $F(x) = \int_{1}^{x} t^2 g(t) dt$, where $g(t) = \int_{1}^{t} f(u) du$, gives $F'(x) = x^2 \int_{1}^{x} f(u) du$. Evaluating this at $x = 1$ causes the integral to span from 1 to 1, resulting in $F'(1) = 1^2 \cdot 0 = 0$, which identifies $x = 1$ as a critical point.

Differentiating $F'(x)$ further using the product rule yields $F''(x) = 2x \int_{1}^{x} f(u) du + x^2 f(x)$. Substituting $x = 1$ into this expression simplifies the calculation because the integral term vanishes, leaving $F''(1) = 2(1)(0) + 1^2 \cdot f(1)$. Given the value $f(1) = 3$, the second derivative evaluates to $F''(1) = 3$. Because the second derivative is positive at this stationary point, the function is concave up, identifying $x = 1$ as a point of local minima.

For three terms to form a geometric progression, the square of the middle term must equal the product of the first and third terms. To determine these terms, we first evaluate the definite integrals involving the fractional part function $\{x\}$ and the greatest integer function $[x]$ over the interval $[0, n]$. The function $\{x\}$ repeats its periodic behavior every unit interval, so the integral $\int_{0}^{n} \{x\} \, dx$ effectively adds up $n$ identical areas, each equal to $\int_{0}^{1} x \, dx$, which simplifies to $n \cdot \frac{1}{2}$ or $\frac{n}{2}$. Similarly, the greatest integer function $[x]$ behaves as a step function that takes constant integer values $0, 1, 2, \dots, n-1$ over successive unit intervals, making its integral $\int_{0}^{n} [x] \, dx$ the sum of these integers, calculated as $\frac{n(n-1)}{2}$.

Substituting these expressions into the geometric progression property $b^2 = ac$ where $a = \frac{n}{2}$, $b = \frac{n(n-1)}{2}$, and $c = 10(n^2 - n)$, we obtain the equation $\left(\frac{n(n-1)}{2}\right)^2 = \frac{n}{2} \cdot 10n(n-1)$. This simplifies to $\frac{n^2(n-1)^2}{4} = 5n^2(n-1)$. Since $n > 1$, we can divide both sides by $n^2(n-1)$, resulting in the simplified equation $\frac{n-1}{4} = 5$. Multiplying by 4 gives $n-1 = 20$, which yields $n = 21$.

The property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$ is essential for simplifying integrands containing a linear factor of $x$, as it allows for the elimination of $x$ when the function is symmetric. By letting the integral be $I$ and replacing $x$ with $2\pi - x$ over the interval $[0, 2\pi]$, we find that the $x$ terms within the numerator cancel, leaving $2I = 2\pi \int_{0}^{2\pi} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} \, dx$, which simplifies the expression to $I = \pi \int_{0}^{2\pi} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} \, dx$.

Because the integrand is periodic with a period of $\pi$, this reduces to $2\pi \int_{0}^{\pi} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} \, dx$. Splitting the range at $\pi/2$ and applying the property $\int_{\pi/2}^{\pi} f(x) \, dx = \int_{0}^{\pi/2} f(\frac{\pi}{2} + x) \, dx$, the integral transforms into $2\pi \int_{0}^{\pi/2} \left( \frac{\sin^8 x}{\sin^8 x + \cos^8 x} + \frac{\cos^8 x}{\sin^8 x + \cos^8 x} \right) \, dx$. This simplifies the integrand to $1$, resulting in $2\pi \int_{0}^{\pi/2} 1 \, dx$, which yields $2\pi \cdot \frac{\pi}{2} = \pi^2$.

The definite integral property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ is highly effective here because the sum of the lower and upper limits is zero. Given:
$I = \int_{-\pi/2}^{\pi/2} \frac{1}{1+e^{\sin x}} dx$
replacing $x$ with $-x$ transforms the expression into $\int_{-\pi/2}^{\pi/2} \frac{1}{1+e^{-\sin x}} dx$. Multiplying the numerator and denominator by $e^{\sin x}$ converts this into $\int_{-\pi/2}^{\pi/2} \frac{e^{\sin x}}{e^{\sin x}+1} dx$. Adding the original integral $I$ to this transformed version results in:
$2I = \int_{-\pi/2}^{\pi/2} \left( \frac{1}{1+e^{\sin x}} + \frac{e^{\sin x}}{e^{\sin x}+1} \right) dx = \int_{-\pi/2}^{\pi/2} 1 dx$
The integral of 1 over the interval $[-\pi/2, \pi/2]$ evaluates to $\pi$. Consequently, $2I = \pi$, which means $I = \frac{\pi}{2}$.

Substituting $x=1$ into the expression reveals a $0/0$ indeterminate form, which necessitates applying L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $x$. Differentiating the integral $\int_0^{(x-1)^2} t \cos(t^2) \, dt$ using the Fundamental Theorem of Calculus yields $2(x-1)^3 \cos((x-1)^4)$, while differentiating the product $(x-1) \sin(x-1)$ via the product rule results in $\sin(x-1) + (x-1) \cos(x-1)$.

This produces the ratio $\frac{2(x-1)^3 \cos((x-1)^4)}{\sin(x-1) + (x-1) \cos(x-1)}$, which remains in an indeterminate state. Dividing both parts of this fraction by $(x-1)$ simplifies it to $\frac{2(x-1)^2 \cos((x-1)^4)}{\frac{\sin(x-1)}{x-1} + \cos(x-1)}$. Evaluating this limit as $x \to 1$ causes $\frac{\sin(x-1)}{x-1}$ to approach $1$ and $\cos(x-1)$ to approach $1$, resulting in a numerator of $0$ and a denominator of $2$, yielding a final value of $0$.

The behavior of a polynomial is defined by its critical points, which occur where the derivative equals zero. Since the critical points of this fourth-degree polynomial are at $-1, 0, 1$, the derivative takes the form $f'(x) = x(x+1)(x-1) = x^3 - x$. Integrating this function leads to $f(x) = \frac{x^4}{4} - \frac{x^2}{2} + C$. To identify the set $T$ where $f(x) = f(0)$, we set the expression equal to $C$, resulting in the equation $\frac{x^4}{4} - \frac{x^2}{2} = 0$. Factoring the expression as $x^2(x^2 - 2) = 0$ reveals the values $0, 0, \sqrt{2},$ and $-\sqrt{2}$. Summing the squares of these values gives $0^2 + 0^2 + (\sqrt{2})^2 + (-\sqrt{2})^2$, which yields a result of $4$.

The definite integral of the polynomial $f(x) = a + bx + cx^2$ from 0 to 1 is calculated by integrating each term, which results in the expression $a + \frac{b}{2} + \frac{c}{3}$. This value represents the total area under the parabola defined by the function within these bounds.

Evaluating $f(x)$ at the specific points used in the options yields $f(0) = a$, $f(1) = a + b + c$, and $f(\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{4}$. Substituting these values into the expression $\frac{1}{6}\{f(0) + f(1) + 4f(\frac{1}{2})\}$ gives $\frac{1}{6}\{a + (a + b + c) + 4(a + \frac{b}{2} + \frac{c}{4})\}$. Simplifying the sum inside the braces results in $6a + 3b + 2c$, and dividing this entire expression by 6 produces $a + \frac{b}{2} + \frac{c}{3}$, confirming that this specific combination of function values is mathematically equivalent to the integral of the quadratic polynomial.

Recognizing the integrand $e^x x^x (2 + \ln x)$ as the derivative of the product $f(x) = e^x x^x$ provides the key to solving the problem immediately. Differentiating this function using the product rule results in $f'(x) = e^x \cdot x^x + e^x \cdot \frac{d}{dx}(x^x)$, and since the derivative of $x^x$ is $x^x(1 + \ln x)$, the full expression expands to $e^x x^x + e^x x^x(1 + \ln x)$. This simplifies directly to $e^x x^x (2 + \ln x)$, which is identical to the given integrand. Because the original function is the exact derivative of $f(x)$, the integral is determined by evaluating $e^x x^x$ at the boundaries 1 and 2. Performing this substitution yields $4e^2 - e$, which factors into the final form $e(4e - 1)$.

The greatest integer function $[3x]$ behaves as a step function, requiring us to partition the integral over the interval $1 \le x \le 2$ at points where $3x$ is an integer, specifically at $x = 4/3$ and $x = 5/3$. Across these segments, the value of $[3x]$ remains constant, which simplifies the absolute value expression to $3-2x$, $4-2x$, and $5-2x$ within their respective sub-intervals. Summing the integrals $\int_{1}^{4/3} (3-2x) dx + \int_{4/3}^{5/3} (4-2x) dx + \int_{5/3}^{2} (5-2x) dx$ results in the addition of $2/9 + 3/9 + 4/9$, providing a final value of 1.0.

Expressing $I_{2}$ by splitting the integrand as $\int_{0}^{1} (1-x^{50})(1-x^{50})^{100} dx$ results in the relation $I_{1} - \int_{0}^{1} x^{50}(1-x^{50})^{100} dx$. Applying integration by parts to the second integral term with $u=x$ and $dv=x^{49}(1-x^{50})^{100} dx$ yields the anti-derivative $v = -\frac{(1-x^{50})^{101}}{5050}$. Because the product $uv$ evaluated at the limits $0$ and $1$ is zero, the integral term simplifies to $\frac{1}{5050} I_{2}$. The equation becomes $I_{2} = I_{1} - \frac{1}{5050} I_{2}$, which rearranges to $\frac{5051}{5050} I_{2} = I_{1}$, identifying the constant $\alpha$ as $\frac{5050}{5051}$.

The fundamental principle applied here is the symmetry property of definite integrals, which states that $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$. By applying this to the given integral $I = \frac{1}{a+b} \int_{a}^{b} x(f(x) + f(x+1)) dx$, we replace $x$ with $a+b-x$ throughout the integrand. Since the provided functional equation $f(a+b+1-x) = f(x)$ implies $f(a+b-x) = f(x+1)$ and $f(a+b-x+1) = f(x)$, the transformed integral becomes $\frac{1}{a+b} \int_{a}^{b} (a+b-x)(f(x+1) + f(x)) dx$.

Summing the original integral and this transformed version effectively eliminates the variable $x$ in the numerator, as $x + (a+b-x) = a+b$. This simplifies the combined expression to $2I = \int_{a}^{b} (f(x) + f(x+1)) dx$. Given that $\int_{a}^{b} f(x+1) dx$ is equivalent to $\int_{a}^{b} f(x) dx$ because they represent the same area under the curve shifted horizontally, we find that $2I = 2 \int_{a}^{b} f(x) dx$, which reduces to $I = \int_{a}^{b} f(x) dx$. Substituting $x = t+1$ into this result shifts the integration limits from $a$ to $b$ into $a-1$ to $b-1$ and transforms the integrand, yielding the final form $\int_{a-1}^{b-1} f(x+1) dx$.

Problems involving the limit of a sum as the number of terms approaches infinity are solved by converting the series into a definite integral using the Riemann sum definition. The series $\sum_{r=1}^{2n} \frac{n}{n^{2+}r^2}$ is simplified by factoring $n^2$ from the denominator, resulting in $\sum_{r=1}^{2n} \frac{1}{n(1 + (r/n)^2)}$.

$\int_{0}^{2} \frac{dx}{1+x^2}$

As $n$ approaches infinity, the discrete fraction $r/n$ becomes the continuous variable $x$ and $1/n$ becomes $dx$, with the bounds ranging from 0 to 2. Evaluating the antiderivative $[\tan^{-1}(x)]_{0}^{2}$ yields $\tan^{-1}(2) - \tan^{-1}(0)$, which results in $\tan^{-1}(2)$.

The absolute value requires splitting the interval of integration at $\pi/2$ because the cosine function changes sign at this point. For the range $[0, \pi/2]$, $\cos x$ is non-negative, so $|\cos x|^3 = \cos^3 x$, whereas for $[\pi/2, \pi]$, $\cos x$ is non-positive, resulting in $|\cos x|^3 = -\cos^3 x$. Applying the trigonometric identity $\cos^3 x = \frac{\cos 3x + 3\cos x}{4}$, the integral becomes $\int_{0}^{\pi/2} \frac{\cos 3x + 3\cos x}{4} \, dx - \int_{\pi/2}^{\pi} \frac{\cos 3x + 3\cos x}{4} \, dx$. Computing this with the antiderivative $\frac{1}{4}(\frac{\sin 3x}{3} + 3\sin x)$ yields $\frac{1}{4}(-\frac{1}{3} + 3) - \frac{1}{4}(0 - (-\frac{1}{3} + 3))$, which simplifies to $\frac{2}{3} + \frac{2}{3}$ for a final total of $4/3$.

The Fundamental Theorem of Calculus establishes that $f'(x)=g(x)$, creating a direct link between the function and its derivative. Applying the substitution $y=t+5$ to the integral $\int_0^x f(t) dt$ updates the integration limits to $[5, x+5]$ and the variable to $dy$. This transformation results in the integral $\int_5^{x+5} f(y-5) dy$, which, based on the properties of the given functions, is equivalent to $-\int_5^{x+5} g(y) dy$. Reversing the limits of integration to remove the negative sign yields $\int_{x+5}^5 g(t) dt$.

Evaluating this integral depends on the property that an odd function integrated over a symmetric interval $[-a, a]$ always sums to zero. Defining the integrand as $h(x) = \ln \left( \frac{2 - x \cos x}{2 + x \cos x} \right)$, substituting $-x$ reveals that $h(-x) = \ln \left( \frac{2 + x \cos x}{2 - x \cos x} \right)$ because $\cos x$ is an even function. This expression simplifies to $h(-x) = - \ln \left( \frac{2 - x \cos x}{2 + x \cos x} \right) = -h(x)$, confirming the function is odd. Consequently, the integral from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$ evaluates to $0$, which is equivalent to $\log_{e} 1$.

The integral $I = \int_{0}^{a} f(x)g(x) dx$ can be rewritten using the definite integral property $\int_{0}^{a} h(x) dx = \int_{0}^{a} h(a-x) dx$ as $I = \int_{0}^{a} f(a-x)g(a-x) dx$. Substituting the given conditions $f(a-x) = f(x)$ and $g(a-x) = 4 - g(x)$ transforms the expression into $I = \int_{0}^{a} f(x)(4 - g(x)) dx$.

Distributing $f(x)$ inside the integral yields $I = 4\int_{0}^{a} f(x) dx - \int_{0}^{a} f(x)g(x) dx$. Since the integral on the far right is the original value $I$, the equation simplifies to $I = 4\int_{0}^{a} f(x) dx - I$, which rearranges to $2I = 4\int_{0}^{a} f(x) dx$. Dividing both sides by 2 gives the final result of $I = 2 \int_{0}^{a} f(x) dx$.

The property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ offers an efficient way to evaluate this integral. By substituting $x$ with $\pi/2 - x$ in the expression $\frac{\sin^3 x}{\sin x + \cos x}$, the sine and cosine functions interchange, leading to the equivalent integral $\int_{0}^{\pi/2} \frac{\cos^3 x}{\cos x + \sin x} dx$. Denoting the original integral as $I$, summing these two forms results in the equation:

$2I = \int_{0}^{\pi/2} \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} dx$

Factorizing the numerator with the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ allows the term $(\sin x + \cos x)$ to cancel out, leaving the integrand as $\sin^2 x - \sin x \cos x + \cos^2 x$. Utilizing the identity $\sin^2 x + \cos^2 x = 1$, the expression simplifies further to $1 - \sin x \cos x$. Calculating the definite integral of this function from $0$ to $\pi/2$ yields $[x - \frac{1}{2}\sin^2 x]_{0}^{\pi/2}$, which evaluates to $(\frac{\pi}{2} - \frac{1}{2}) - (0 - 0)$. Dividing this result by two confirms the value is $\frac{\pi-1}{4}$.

Transforming the integrand $\sec^{2/3} x \csc^{4/3} x$ into $\sec^2 x \tan^{-4/3} x$ allows for the substitution $u = \tan x$, where $du = \sec^2 x \, dx$ and the limits change from $x \in [\pi/6, \pi/3]$ to $u \in [1/\sqrt{3}, \sqrt{3}]$. Evaluating the resulting integral $\int_{1/\sqrt{3}}^{\sqrt{3}} u^{-4/3} \, du$ gives the antiderivative $-3u^{-1/3}$, which at these boundaries becomes $-3[(\sqrt{3})^{-1/3} - (1/\sqrt{3})^{-1/3}]$. Simplifying these terms using exponent laws produces $3^{7/6}-3^{5/6}$.

Recognizing this expression as a Riemann sum allows us to convert the limit of a summation into a definite integral. We rewrite the denominator $n^{4/3}$ as $n \cdot n^{1/3}$ and distribute the $n^{1/3}$ across the summation terms to obtain $\frac{1}{n} \sum_{r=1}^{n} \left(\frac{n+r}{n}\right)^{1/3}$, which simplifies to $\frac{1}{n} \sum_{r=1}^{n} \left(1 + \frac{r}{n}\right)^{1/3}$. As $n$ approaches infinity, the sum converges to the integral $\int_{0}^{1} (1+x)^{1/3} \, dx$, where the term $\frac{r}{n}$ transitions into the continuous variable $x$. Computing this integral results in $\left[ \frac{3}{4} (1+x)^{4/3} \right]_{0}^{1}$, which evaluates to $\frac{3}{4} (2^{4/3} - 1)$, and ultimately leads to $\frac{3}{4} (2)^{4/3} - \frac{3}{4}$.